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Dedicated to Professor Neil Trudinger on the occasion of his 80th birthday.
We consider the fourth-order total variation flow equation in Rn of the form
ut=−Δdiv∇u|∇u|. | (1.1) |
We aim to give explicit description of its solutions emanating from piecewise constant radial data. However, it turns out that the definition of a solution is itself non-trivial since −Δ does not have a bounded inverse on L2(Rn). Our first goal is thus to provide a rigorous definition of a solution. Our second goal is to find explicit formula for the solution to (1.1) when the initial datum u(0,x)=u0(x) is the characteristic function of a ball or an annulus. In other words,
u0=a01BR0oru0=a01AR10R00a0∈R, |
where 1K is the characteristic function of a set K⊂Rn, i.e.,
1K(x)={1,x∈K0,x∈Rn∖K. |
Here BR denotes the open ball of radius R centered at 0∈Rn and AR1R0 denotes the annulus defined by AR1R0=BR1∖¯BR0. Our major concern is whether or not the solution remains a characteristic function throughout the evolution. For example, in the case u0=a01BR0, whether or not the solution u of (1.1) is of the form
u(t,x)=a(t)1BR(t) |
with a function a=a(t). In other words, we are asking whether the speed ut on the ball BR(t) and on its complement are constant in the spatial variable. As in the second-order problem [3] (see also [17]), this leads to the notion of calibrability of a set. In the case of the second-order problem ut=div(∇u/|∇u|), a ball and its complement are always calibrable and R(t)≡R0, i.e., the ball does not expand nor shrink [3]. In our problem, R(t) may be non-constant.
We first note that the definition of a solution itself is non-trivial. The fourth-order total variation flow has been mainly studied in the periodic setting [12,14] or in a bounded domain with some boundary conditions [15]. Formally, it is a gradient flow of the total variation functional
TV(u):=∫Ω|∇u| |
with respect to the inner product
(u,v)−1=∫Ωu(−Δ)−1v |
when Ω is a domain in Rn or a flat torus Tn. In the periodic setting, i.e., Ω=Tn as in [12,14], it is interpreted as a gradient flow in H−1av which is the dual space of H1av, the space of average-free H1 functions equipped with the inner product
(u,v)1=∫Ω∇u⋅∇v. |
For the homogeneous Dirichlet boundary condition with bounded Ω, H−1av is replaced by D−1, the dual space of D10=D10(Ω), which is the completion of C∞c(Ω) in the norm associated with the inner product (u,v)1; here C∞c(Ω) denotes the space of all smooth functions compactly supported in Ω. By the Poincaré inequality, both H1av and D10(Ω) can be regarded as subspaces of L2(Ω). However, if Ω equals Rn, the situation is more involved. If n≥3, D10(Rn) is continuously and densely embedded in L2∗(Rn), where p∗=np/(n−p) so that 2∗=2n/(n−2), by the Sobolev inequality. In fact,
D10(Rn)=D1(Rn)∩L2∗(Rn),D1(Rn)={u∈L1loc(Rn)|∇u∈L2(Rn)} |
see e.g., [11]. On the other hand, if n≤2, D10 is isometrically identified with the quotient space ˙D1(Rn):=D1(Rn)/R, when D1(Rn) is equipped with inner product (u,v)1 [11]. Thus, we need to be careful when n≤2 because an element of D10(Rn) is determined only up to a constant. In any case, D10(Rn) is a Hilbert space with the scalar product
(u,v)D10(Rn)=∫Rn∇u⋅∇v. |
Therefore, we can identify D10(Rn) with its dual space by means of the isometry
−Δ:u↦(u,⋅)D10(Rn). |
On the other hand, let us define a subspace ˜D−1(Rn)⊂D10(Rn)′ by
˜D−1(Rn)={w↦∫Rnuw:u∈C∞c(Rn)}if n≥3, |
˜D−1(Rn)={w↦∫Rnuw:u∈C∞c,av(Rn)}if n=1 or n=2, |
where
C∞c,av(Rn)={u∈C∞c(Rn):∫Rnu=0}. |
Then the closure D−1(Rn) of ˜D−1(Rn) coincides with D10(Rn)′ [11]. Note that the restriction to C∞c,av(Rn) in the definition of ˜D−1(Rn) in n=1,2 is necessary for the functionals to be well-posed on D1(Rn)/R. In any case, since (by definition) the space of test functions D(Rn) is continuously and densely embedded in D10(Rn), we also have a continuous embedding D−1(Rn)=D10(Rn)′⊂D′(Rn). Throughout the paper, we will often drop (Rn) in the notation for spaces of functions on Rn, e.g., D−1=D−1(Rn).
In the first step, we give a rigorous definition of the total variation functional TV on D−1. Then we calculate the subdifferential of TV in D−1 space. Since it is a homogeneous functional, we are able to apply a duality method [3] to characterize the subdifferential, provided that TV is well approximated by nice functions in D−1. We know that C∞c,av(Rn) is dense in D−1 for n≤2; see e.g., [11]. However, it is not immediately clear whether TV is simultaneously approximable. Fortunately, it turns out that for any w∈D−1, there is a sequence wk∈C∞c,av(Rn) which converges to w in D−1 and TV(wk)→TV(w) as k→∞. This approximation part is relatively involved since we have to use an efficient cut-off function. Using the approximation, we are able to characterize the subdifferential ∂D−1TV of TV in D−1 by adapting the argument in [3]. Namely, we have
∂D−1TV(u)={v=ΔdivZ|Z∈L∞(Rn), |Z|≤1, −⟨u,divZ⟩=TV(u)}, |
where ⟨, ⟩ denotes the canonical paring of D−1 and D10. A vector field Z corresponding to an element of the subdifferential is often called a Cahn-Hoffman vector field. The equation (1.1) should be interpreted as the gradient flow of TV in D−1, i.e.,
ut∈−∂D−1TV(u), | (1.2) |
and its unique solvability for any initial datum u0∈D−1 is guaranteed by the classical theory of maximal monotone operators ([6,24]). By our characterization of the subdifferential, we are able to give a more explicit definition of a solution which is consistent with that proposed in [15]. Namely, for u0∈D−1 with TV(u0)<∞, a function u∈C([0,T[,D−1) is a solution to (1.2) with u(0)=u0 if and only if there exists Z∈L∞(]0,T[×Rn) satisfying divZ∈L2(0,T;D10(Rn)) such that
ut=−ΔdivZinD−1(Rn) | (1.3) |
|Z(t,x)|≤1for a.e. x∈Rn | (1.4) |
⟨u,divZ⟩=−TV(u) | (1.5) |
for a.e. t∈]0,T[. This is convenient for calculating explicit solutions.
Unfortunately, in n≤2, for a compactly supported square integrable function u0, we know that u0∈D−1 if and only if u0 is average-free, i.e., ∫Rnu0=0 (see Lemma 17). Thus, the characteristic function of any bounded, measurable set of positive measure does not belong to D−1. We have to extend a class of initial data u0 such that u0=ψ+w0 with w0∈D−1 while ψ is a fixed compactly supported L2 function. We consider a gradient flow ut∈−∂D−1TV(u) in the affine space ψ+D−1. Since ∂D−1 is a directional partial derivative in the direction of D−1, it is more convenient to consider solutions to evolutionary variational inequality
12ddt‖u(t)−g‖2D−1≤TV(g)−TV(u(t))for a.e. t>0 | (1.6) |
for any g∈ψ+D−1 [2]. In the case ψ=0, it is easy to show that the evolutionary variational inequality is equivalent to (1.2). Indeed, by definition of the subdifferential, (1.2) is equivalent to
(−ut,g−u(t))D−1≤TV(g)−TV(u(t)) |
for any g∈D−1. The left-hand side equals (d/dt)(‖u−g‖2/2). Thus, the equivalence follows if ψ=0. From now on we assume that ∫Rnψ≠0.
It is easy to check that there is at most one solution to the evolutionary variational inequality (1.6). The solution u is constructed by solving
wt∈−∂D−1TV(w+ψ)withw(0)=w0=u0−ψ |
and setting u=w+ψ. Characterization of the (directional) subdifferential is more involved since w↦TV(w+ψ) is no more positively one-homogeneous. We identify the one dimensional space {cψ|c∈R} with R and consider the Hilbert space E−1 defined as the orthogonal sum D−1⊕R. We calculate the subdifferential by the duality method since TV is now positively one-homogeneous on E−1. We then project this subdifferential onto D−1 to get a characterization of a (directional) subdifferential ∂D−1TV. We end up with a characterization of solution to (1.2) similar to (1.3)–(1.5), with (1.5) adjusted in a suitable way. If we also denote E10=D1 in n≤2, E10=D10, E−1=D−1 in n≥3 and
⟨u,v⟩E={⟨u,v⟩ if n≥3,⟨w,[v]⟩+c∫ψv, where u=w+cψ, w∈D−1 if n≤2, | (1.7) |
for u∈E−1, v∈E10, we end up with the following definition of solution.
Definition 1. Assume that u0∈E−1. We say that u∈C([0,∞[,E−1) with ut∈L2loc(]0,∞[,D−1) is a solution to (1.1) with initial datum u0 if there exists Z∈L∞(]0,∞[×Rn) with divZ(t,⋅)∈E10 for a.e. t>0 such that
ut=−ΔdivZinD−1(Rn) | (1.8) |
|Z(t,x)|≤1fora.e.x∈Rn | (1.9) |
−⟨u,divZ⟩E=TV(u) | (1.10) |
for a.e. t>0.
and associated well-posedness result.
Theorem 2. Let u0∈E−1. There exists a unique solution to (1.1) with initial datum u0.
Our next problem is whether or not the speed of a characteristic function of a set is spatially constant inside and outside of the set. By the general theory ([6,24]), the speed is determined by the minimal section (canonical restriction) ∂0D−1TV of ∂D−1TV. In other words, ∂0D−1TV(u)=v0 minimizes ‖v‖D−1 for v∈∂D−1TV(u), i.e.,
∂0D−1TV(u):=argmin{‖v‖D−1|v∈∂D−1TV(u)}. |
To motivate the notion of calibrability, we consider a smooth function u such that
¯U={x∈Rn|u(x)=0}, |
where U is a smooth open set. Outside ¯U, we assume that ∇u≠0. To fix the idea, we assume that ∂U has negative signature (orientation) in the sense that u<0 outside ¯U. By our specification of u, we see that
∂0D−1TV(u)=argmin{‖divZ‖D10||Z|≤1 in U, Z=∇u/|∇u| in ¯Uc, divZ∈D10}. |
Since divZ is locally integrable, Z⋅ν does not jump across ∂U, where ν is the exterior unit normal of ∂U. In this case,
Z⋅ν=Z⋅∇u/|∇u|=−1on ∂U. | (1.11) |
Since ∇divZ does not have a singular part, divZ does not jump across ∂U. In this case,
divZ=−divνon ∂U. | (1.12) |
However, v=ΔdivZ may have a non-zero singular part concentrated on ∂U even if v=v0, i.e., v is the minimizer. This phenomenon is observed in [12,20,21] in a one-dimensional periodic setting. Different from the second-order problem, this causes expansion or shrinking of the ball when the solution u is of the form u(t,x)=a(t)1BR(t)(x). If u>0 outside U, the minus in (1.11) and (1.12) should be replaced by the plus.
If ∂0D−1TV(u) is constant on BR(t) and (¯BR(t))c, this property is preserved under the evolution, which leads us to definition of calibrability. Note that the value of ∂0D−1TV(u) on U is determined by U and its signature does not depend on particular value of u. We say that U (with negative signature) is calibrable if there exists Z0∈L∞(U,Rn) such that ∇divZ0∈L2(U,Rn), Z0 satisfies (1.11), (1.12), |Z0|≤1 a.e. in U and ΔdivZ0 is constant on U. We call any such Z0 a calibration for U.
Recall that in the case of the second-order problem, we say that U (with negative signature) is calibrable if there exists ˜Z0∈L∞(U,Rn) satisfying (1.11), |˜Z0|≤1 a.e. in U and −div˜Z0 is a constant function on U. This is formally equivalent to −calibrability in [3,4]. It can be shown that ˜Z0 is a calibration for U if and only if
˜Z0∈argmin{‖divZ‖L2(U)|z satisfies (1.11) and |Z|≤1 a.e. in U} |
and −div˜Z0 is a constant function on U, which is the definition of calibrability in [17].
Going back to our fourth-order problem, if Z0 is a calibration for U, then w=divZ0 must satisfy
−Δw=λin U | (1.13) |
w=−divνon ∂U | (1.14) |
with some constant λ. If U is bounded, λ is determined by (1.11) since
∫UwdLn=∫UdivZ1dLn=∫∂UZ1⋅νdHn−1=−Hn−1(∂U), | (1.15) |
where Hn−1 denotes the n−1 dimensional Hausdorff and Ln denotes the Lebesgue measure in Rn. Using this fact, in Section 5 we prove that if Z0 is a calibration for a bounded U, then
Z0∈argmin{‖∇divZ‖L2(U)|Z satisfies (1.11), (1.12) and |Z|≤1 a.e. in U}. | (1.16) |
Moreover, we obtain an "explicit" formula for the constant λ in terms of the Saint-Venant problem in U.
In the radially symmetric setting, it is not difficult to show that Z0 in (1.16) can be chosen in the form z(|x|)x|x|. Indeed, if Z0 is belongs to the set of minimizers (1.16), then its rotational average ¯Z0 belongs to (1.16) as well, because averaging preserves (1.11), (1.12) and the inequality |Z|≤1. Since the angular part of ¯Z0 does not contribute to the divergence, it is possible to delete this part (Lemma 31). We thus conclude that there is an element of (1.16) of form Z(x)=z(|x|)x|x|. Thus, the Eq (1.13) can be written as the third-order ODE of the form
−r1−n(rn−1(r1−n(rn−1z)′)′)′=λ | (1.17) |
since divZ=r1−n(rn−1z)′. If U is BR with negative signature, condition (1.11) implies
z(R)=−1. | (1.18) |
Since divZ=z′+(n−1)z/r, condition (1.12) implies that
z′(R)=0. | (1.19) |
Solving (1.17) under the assumption that z is smooth near zero under conditions (1.18), (1.19), we eventually get a unique solution (1.17)–(1.19) of the form
z(r)=12(rR)3−32rR,λ=−n(n+2)R3 |
for all n≥1. It is easy to see that Z(x)=z(|x|)x|x| satisfies the constraint |Z|≤1 in BR. We conclude that all balls are calibrable. More careful argument is necessary, but we are able to discuss calibrability of an annulus as well as a complement of a ball.
Theorem 3. (i) All balls are calibrable for all n≥1.
(ii) All complement of balls are calibrable except n=2.
(iii) If n=2, all complement of balls are not calibrable.
(iv) All annuli (with definite signature) are calibrable except in n=2.
(v) For n=2, there is Q∗>1 such that an annulus (with definite signature) is calibrable if and only if the ratio of the exterior radius over the interior radius is smaller than or equal to Q∗. In other words, AR1R0 is calibrable if and only if R1/R0≤Q∗.
Theorem 3(v) is consistent with (iii) since R1→∞ implies AR1R0 converges to ¯BR0c, a complement of the closure of the ball BR0. Note that in the case of an annulus, there is a possibility we take a signature which is different on the exterior boundary ∂BR1 and the interior boundary ∂BR0. We also study such indefinite cases.
We now calculate an explicit solution of (1.1) starting from u0=a01BR0. We first discuss the case n≠2. Since a ball and its complement is calibrable, the solution is of the form
u(t,x)=a(t)1BR(t). | (1.20) |
We take the (radial) calibration Zin in BR(t) and Zout in Rn∖¯BR(t) and set
Z(x,t)={Zin(x),x∈BR(t)Zout(x),x∈Rn∖¯BR(t). |
Here Zout(x)=zout(|x|)x|x| can be calculated as
zout(r)=−n−12(rR)3+n−32(rR)1−n |
while, as we already discussed, zin for Zin(x)=zin(|x|)x|x| is of the form
zin(r)=12(rR)3−32rR. |
This Z satisfies (1.9) and (1.10), and moreover divZ∈D10 for any t>0. Moreover, divZ is continuous across ∂BR(t). However, ∇divZ may jump across ∂BR(t). Actually,
−ΔdivZ=λ1BR(t)+ν⋅(∇divZin−∇divZout)δ∂BR(t), |
where δΓ(φ)=∫ΓφdHn−1 or \delta_\Gamma = \mathcal{H}^{n-1}\, \, ∟\, \Gamma for a hypersurface \Gamma and \nu is the exterior unit normal of \partial B_{R(t)} , i.e., \nu = x/R(t) . Here \lambda = -\frac{n(n+2)}{R^3} . By a direct calculation, the quantity \nu\cdot (\nabla\operatorname{div}Z_{in}-\nabla\operatorname{div}Z_{out}) = -\frac{n(n-4)}{R^2} . Since u_t = -\Delta\operatorname{div}Z , by
\partial_t\left(a\mathbf{1}_{B_R}\right) = \frac{\mathrm{d}a}{\mathrm{d}t}\mathbf{1}_{B_R} + a \frac{\mathrm{d}R}{\mathrm{d}t}\delta_{\partial B_R}, |
we conclude that
\frac{\mathrm{d}a}{\mathrm{d}t} = - \frac{n(n+2)}{R^3}, \quad \frac{\mathrm{d}R}{\mathrm{d}t} = - \frac{n(n-4)}{aR^2}. |
Since
\frac{\mathrm{d}}{\mathrm{d}t} (aR^3) = - n(n+2) - 3n(n-4) = -n(4n-10), |
an explicit form of a solution is given as
a(t) = a_0 \left(1 - \frac{n(4n-10)}{a_0 R^3_0}t\right)^{\frac{n+2}{4n-10}}, \quad R(t) = R_0 \left(1 - \frac{n(4n-10)}{a_0 R^3_0}t\right)^{\frac{n-4}{4n-10}}. |
As noticed earlier, in the case n = 2 , the complement of the disk is not calibrable. If u is a radially strictly decreasing function outside B_R , we expect Z_{out}(x) = -x/|x| for |x| > R(t) . In [15], it is proposed that a solution u to (1.1) must satisfy
u_t = -\Delta\operatorname{div}Z_{out}. |
Since \operatorname{div}Z_{out} = -(n-1)/|x|^2 and \nabla\operatorname{div}Z_{out} = \frac{(n-1)x}{|x|^3} , this implies
\begin{equation} u_t(t, x) = -\frac{(n-1)(n-3)}{|x|^3}, \quad x \in \left(\overline{B_{R(t)}}\right)^c = \mathbb{R}^n\backslash\overline{B_{R(t)}}. \end{equation} | (1.21) |
In the case n = 2 , \nabla\operatorname{div}Z_{out}\in L^2\left(\left(\overline{B_{R(t)}}\right)^c\right) so Z_{out} is a Cahn-Hoffman vector field.
If we start with u_0 = a_0\mathbf{1}_{B_{R_0}} with a_0 > 0 for n = 2 , the expected form of a solution is
\begin{equation} u(t, x) = a(t)\mathbf{1}_{B_{R(t)}}(x) + \frac{t}{|x|^3}\mathbf{1}_{\overline{B_{R(t)}}^c}(x), \end{equation} | (1.22) |
where
\begin{equation} \frac{\mathrm{d}a}{\mathrm{d}t} = -\frac{2\cdot 4}{R^3}, \quad \left( a(t) - \frac{t}{R(t)^3} \right) \frac{\mathrm{d}R}{\mathrm{d}t} = \frac{2\cdot 2}{R^2}. \end{equation} | (1.23) |
Analyzing this ODE system, we can deduce qualitative properties of the solution. Summing up our results yields.
Theorem 4. Let u_0 = a_0\mathbf{1}_{B_{R_0}} with a_0 > 0 .
If n \geq 3 , then the solution u to (1.1) with initial datum u_0 is of the form
u(t, x) = a(t)\mathbf{1}_{B_R(t)} \quad\mathit{\text{for}}\quad t < t_* = a_0 R^3_0\bigm/\left(n(4n-10)\right) |
and u(t, x)\equiv0 for t\geq t_* . (The time t_* is called the extinction time.) Moreover, a(t) is decreasing and a(t)\to0 as t\uparrow t_* .
(i) R(t) is increasing and R(t)\to\infty as t\uparrow t_* for n = 3 .
(ii) R(t) = R_0 for n = 4 .
(iii) R(t) is decreasing and R(t)\to0 as t\uparrow t_* for n\geq5 .
If n = 2 , then the solution is not a characteristic function for t > 0 . It is of the form (1.21) and moves by (1.23). In particular, there is no extinction time, R(t) is increasing and a(t) is decreasing. Moreover, R(t)\to\infty and a(t)\to0 as t\to\infty . The gap a(t)-\frac{t}{R(t)^3} is always positive.
If n = 1 , then the solution is of the form u(t, x) = a(t)\mathbf{1}_{B_R(t)} for t > 0 . Moreover, R(t) is increasing and a(t) is decreasing with R(t)\to\infty and a(t)\to0 as t\to\infty .
We note that the infinite extinction time observed in n \leq 2 is related to the fact that 0 is not an element of the affine space u_0+D^{-1} where the flow lives if \int u_0 \neq 0 . In [14], finite time extinction for solution to (1.1) is proved in a periodic setting for average zero initial data when the space dimension n\leq4 Our result is unrelated to their result because we consider (1.1) in \mathbb{R}^n .
The formula (1.21) does not give a solution to (1.1) when n\geq4 since \nabla\operatorname{div}Z_{out} does not belong to L^2\left(\left(\overline{B_{R(t)}}\right)^c\right) . In the case n = 3 , this formula is consistent with our definition. If we consider u_0 strictly radially decreasing for |x| > R_0 and u_0(x) = u_* for |x|\leq R_0 , then u_0 does not belong to the domain of \partial_{D^{-1}}TV for n\geq4 . In other words, there is no Cahn-Hoffman vector field.
These results contrast with the second-order total variation flow
u_t = \operatorname{div}\left(\nabla u/|\nabla u|\right). |
In the second-order problem, a ball and an annulus are always calibrable with their complements, see e.g., [3] or [17, Section 5]. Furthermore, u_t(t, \cdot) is a locally integrable function without singular part for t > 0 . Thus, for example, the solution starting from u_0 = a_0\mathbf{1}_{B_{R_0}} ( a_0 > 0 ) must be u(t, x) = a(t)\mathbf{1}_{B_{R_0}} with a(t) = -\lambda t+a_0 , where \lambda is the Cheeger ratio, i.e., \lambda = \mathcal{H}^{n-1}(\partial B_{R_0})/\mathcal{L}^n(B_{R_0}) . In particular, the extinction time t_* equals t_* = a_0/\lambda .
We conclude this paper by deriving a system of ODEs prescribing the solution in the case when the initial datum is a piecewise constant, radially symmetric function, which we call a stack. To be precise, we say that w \in E^{-1} is a stack if it is of the form
w = a^0 \mathbf{1}_{B_{R^0}} + a^1 \mathbf{1}_{A_{R^0}^{R^1}} +\ldots + a^{N-1} \mathbf{1}_{A_{R^{N-2}}^{R^{N-1}}} + a^N \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}}}, |
0 < R^0 < R^1 < \ldots < R^{N-1} , a^k \in \mathbb{R} . In particular, we obtain
Theorem 5. Let n \neq2 and let u_0 be a stack. If u is the solution to (1.1), then u(t, \cdot) is a stack for t > 0 .
In the case n = 2 , this result is no longer true, as evidenced by Theorem 4. However the solution can still be prescribed by a finite system of ODEs.
A total variation flow type equation
\begin{equation} w_t = -\Delta \left( \operatorname{div}\left(\nabla w /|\nabla w|\right)+\beta\operatorname{div}\left(\nabla w|\nabla w|\right) \right) \end{equation} | (1.24) |
was introduced by [31] to describe the height of crystal surface moved by relaxation dynamics below the roughening temperature, where \beta > 0 . For this equation, characterization of the subdifferential of the corresponding energy was given by Y. Kashima in periodic setting [20,21] and under Dirichlet condition on a bounded domain [21]. The speed of a facet (a flat part of the graph) is calculated for n = 1 in [20] and for a ball with the Dirichlet condition under radial symmetry [21]. Different from the second-order problem, the speed of a facet is determined not only by the shape of facet. Also it has been already observed in [20], that the minimal section may have a delta part although the behavior of the corresponding solution was not studied there. A numerical computation was given in [23]. The Eq (1.24) was derived as a continuum limit of models describing motion of steps on crystal surface as discussed in [27], where numerical simulation was given; see also [22].
In [7], a crystalline diffusion flow was proposed and calculated numerically. In a special case, it is of the form w_t = -\partial^2_x\left(W'(w_x)\right) , where W is a piecewise linear convex function, when the curve is given as the graph of a function. This equation was analyzed in [13] in a class of piecewise linear (in space) solutions.
Fourth-order equations of type (1.1) were proposed for image denoising as an improvement over the second-order total variation flow. For example, the equation
w_t = -\Delta\operatorname{div}\left( \nabla w/|\nabla w| \right) + \lambda(f-w), |
where f is an original image which is given and \lambda > 0 , corresponds to the Osher-Solé-Vese model [28]. The well-posedness of this equation was proved by using the Galerkin method by [8].
For (1.1), an extinction time estimate was given in [14] for n = 1, 2, 3, 4 in the periodic setting. It was extended to the Dirichlet problem in a bounded domain by [15]. In the review paper [12], it was proved that the solution u of (1.1) in n = 1 may become discontinuous instantaneously even if the initial datum is Lipschitz continuous, because the speed may have a delta part.
There are a few numerical studies for (1.1) in the periodic setting. A duality-based numerical scheme which applies the forward-backward splitting has been proposed in [16]. A split Bregman method was adjusted to (1.1) and also (1.24) in [18]. In these methods, the singularity of the equation at \nabla u = 0 is not regularized. However, all above studies deal with either periodic, Dirichlet or Neumann boundary condition for a bounded domain. It has never been rigorously studied in \mathbb{R}^n , although in [15] there are some preliminary calculations for radial solution in \mathbb{R}^n .
This paper is organized as follows. In Section 2, we discuss basic properties of the total variation on D^{-1} , notably we show strict density of C^\infty_{c, av} . In Section 3, we give a rigorous definition of a solution to (1.1) and obtain a verifiable characterization of solutions. In Section 4, we extend the results of the previous section to include initial data with non-zero average in n = 1, 2 . In Section 5, we introduce the notion of calibrability. In Section 6, we discuss calibrability of rotationally symmetric sets in \mathbb{R}^n . In Section 7, we study solutions emanating from piecewise constant, radially symmetric data.
In this section, we give a rigorous definition of the total variation TV on D^{-1} and relate it to the usual total variation defined on L^1_{loc} . The main tool that we use here as well as in the following section is an approximation lemma, which for a given w \in D^{-1} produces a sequence of nice functions w_k \in D^{-1} that converges to w in D^{-1} and TV(w_k) \to TV(w) .
Let us denote
X_1 = \left\{ \psi \in C^\infty_c(\mathbb{R}^n, \mathbb{R}^n), \ \|\psi\|_{L^\infty(\mathbb{R}^n, \mathbb{R}^n)} \leq 1\right\}. |
We define TV \colon D^{-1}(\mathbb{R}^n) \to [0, \infty] by
TV(u) = \sup\limits_{\psi \in X_1} \langle u, \operatorname{div}\, \psi\rangle. |
Let us compare this definition with the usual total variation, which we denote here by \overline{TV}\colon L^1_{loc}(\mathbb{R}^n) \to [0, \infty] , defined by
\overline{TV}(u) = \sup\limits_{\psi \in X_1} \int_{\mathbb{R}^n} u \, \operatorname{div}\, \psi. |
First of all, as in the case \overline{TV} , we easily check that TV is lower semicontinuous with respect to the weak-* (and, a fortiori, strong) convergence in D^{-1}(\mathbb{R}^n) . Indeed, if v_k \overset{*}{\mathop{\rightharpoonup }} v in D^{-1}(\mathbb{R}^n) ,
\begin{equation*} TV(v) = \sup\limits_{\psi \in X_1} \{\langle v, \operatorname{div}\, \psi \rangle\} = \sup\limits_{\psi \in X_1} \liminf\limits_{k\to \infty} \{\langle v_k, \operatorname{div}\, \psi \rangle\} \leq \liminf\limits_{k\to \infty} \sup\limits_{\psi \in X_1} \{\langle v_k, \operatorname{div}\, \psi \rangle\} = \liminf\limits_{k\to \infty} TV(v_k). \end{equation*} |
In fact, we have
Lemma 6. We have D(TV) \subset L^1_{loc} , and so D(TV) \subset D(\overline{TV}) with TV and \overline{TV} coinciding on D(TV) . In particular, if n \geq 2 , D(TV) \subset L^{1^*}(\mathbb{R}^n) . If n = 1 ,
D(TV) \subset L^\infty_0(\mathbb{R}) = \left\{w \in L^\infty(\mathbb{R}) \colon \underset{x \rightarrow \pm \infty}{\operatorname{ess} \lim } w(x) = 0 \right\}. |
The proof of this fact is a consequence of the lemma below and we postpone it.
Lemma 7. For any w \in D^{-1}(\mathbb{R}^n) there exists a sequence w_k \in C^\infty_{c, av}(\mathbb{R}^n) such that
w_k \to w \;\mathit{\text{in}}\; D^{-1}(\mathbb{R}^n) |
and
TV(w_k) \to TV(w). |
To prove it, we will use a special choice of cut-off function and associated variant of the Sobolev-Poincaré inequality. For R > 0 , let us denote by \vartheta_R the element of minimal norm in D^1_0(\mathbb{R}^n) among those w \in D^1_0(\mathbb{R}^n) that satisfy w(x) = 1 if |x|\leq \frac{R}{2} , w(x) = 0 if |x| \geq R . It is an easy exercise to show that for \frac{R}{2} \leq |x| \leq R
\begin{equation*} \label{vartheta_form} \vartheta_R(x) = \left(2^{n-2}-1\right)^{-1}\left(\left(\frac{|x|}{R}\right)^{2-n} -1 \right) \text{ if } n \neq 2, \quad \vartheta_R(x) = \frac{\log\frac{R}{|x|}}{\log 2} \text{ if } n = 2. \end{equation*} |
In either case,
\begin{equation} \nabla \vartheta_R(x) = C_n \frac{|x|^{-n}x}{R^{2-n}} \text{ if } \frac{R}{2} \leq |x| \leq R. \end{equation} | (2.1) |
Lemma 8. If p \in [1, n[ and q \in [1, p^*] , then for all w \in C^1(\mathbb{R}^n) , R > 0 there holds
\begin{equation} \left\| w - \frac{\int \vartheta_R w}{\int \vartheta_R}\right\|_{L^{q}(B_R)} \leq C R^{1+\frac{n}{q} - \frac{n}{p}} \|\nabla w\|_{L^p(B_R)} \end{equation} | (2.2) |
with C = C(n, p) and
\begin{equation} \|\nabla \vartheta_R\|_{L^p(\mathbb{R}^n)} = C R^{-\frac{1}{p}(n-1)(2-p)} \end{equation} | (2.3) |
with a different C = C(n, p) .
Proof. Let v \in C^1(\mathbb{R}^n) . Following the proof of the standard Poincaré inequality by contradiction using Rellich-Kondrachov theorem, we obtain
\left\| v - \frac{\int \vartheta_1 v}{\int \vartheta_1}\right\|_{L^{p}(B_1)} \leq C \|\nabla v\|_{L^p(B_1)}. |
Applying the Sobolev inequality in B_1 to the function v - \frac{\int \vartheta_1 v}{\int \vartheta_1} , we upgrade this to
\begin{equation} \left\| v - \frac{\int \vartheta_1 v}{\int \vartheta_1}\right\|_{L^{q}(B_1)} \leq C \|\nabla v\|_{L^p(B_1)} \end{equation} | (2.4) |
Next, let v(x) = w(Rx) for a given w \in C^1(\mathbb{R}^n) . We observe that
\vartheta_1(x) = \vartheta_R(Rx) \quad \text{for } x \in \mathbb{R}^n |
and so, by a change of variables x = y/R ,
\int \vartheta_1 = \frac{1}{R^n}\int \vartheta_R, \quad \int \vartheta_1 v = \frac{1}{R^n}\int \vartheta_R w. |
Applying the same change of variables to both sides of (2.4) we conclude the proof of (2.2).
The proof of (2.3) is a matter of direct calculation.
Let us now return to the proof of the approximation lemma.
Proof of Lemma 7. Given w \in D^{-1} , let
w_{\varepsilon, R} = \left(\varrho_\varepsilon * w - \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R}\right) \vartheta_R. |
Equivalently, for \varphi \in D_0^1(\mathbb{R}^n) ,
\langle w_{\varepsilon, R}, \varphi \rangle = \langle w, \varrho_\varepsilon * \left(\left(\varphi - \frac{\int \vartheta_R \varphi}{\int \vartheta_R}\right) \vartheta_R\right)\rangle. |
Denoting \widetilde{w} = (-\Delta)^{-1} w ,
\begin{eqnarray} \langle w_{\varepsilon, R} - w, \varphi \rangle = \int \nabla \varrho_\varepsilon * \widetilde{w} \cdot \nabla \left(\left(\varphi - \frac{\int \vartheta_R \varphi}{\int \vartheta_R}\right) \vartheta_R - \varphi\right) + \int \left(\nabla \varrho_\varepsilon * \widetilde{w} - \nabla \widetilde{w}\right)\cdot \nabla \varphi\\ = \int \nabla \varrho_\varepsilon * \widetilde{w} \cdot (\vartheta_R - 1) \nabla \varphi + \int \nabla \varrho_\varepsilon * \widetilde{w} \cdot \left(\varphi - \frac{\int \vartheta_R \varphi}{\int \vartheta_R}\right) \nabla \vartheta_R + \int \left(\nabla \varrho_\varepsilon * \widetilde{w} - \nabla \widetilde{w}\right)\cdot \nabla\varphi. \end{eqnarray} |
We estimate the second term on the r. h. s. using the Poincaré inequality from Lemma 8, taking into account that the support of the integrand is contained in \overline{A}_R , where A_R = B_R \setminus \overline{B}_{R/2} ,
\begin{eqnarray} \left|\int \nabla \varrho_\varepsilon * \widetilde{w} \cdot \left(\varphi - \frac{\int \vartheta_R \varphi}{\int \vartheta_R}\right) \nabla \vartheta_R\right| \leq C \|\nabla \varrho_\varepsilon * \widetilde{w} \, \mathbf 1_{A_R}\|_{L^2(\mathbb{R}^n)} \left\|\varphi - \frac{\int \vartheta_R \varphi}{\int \vartheta_R}\right\|_{L^2(\mathbb{R}^n)} \|\nabla \vartheta_R\|_{L^\infty(\mathbb{R}^n)} \\ \leq C \|\nabla \varrho_\varepsilon * \widetilde{w} \, \mathbf 1_{A_R}\|_{L^2(\mathbb{R}^n)} \|\nabla \varphi\|_{L^2(\mathbb{R}^n)}. \end{eqnarray} |
Thus,
\begin{eqnarray} \|w_{\varepsilon, R} - w\|_{D^{-1}(\mathbb{R}^n)} = \sup\limits_{\|\varphi\|_{D^1_0(\mathbb{R}^n) }\leq 1} \langle w_{\varepsilon, R} - w, \varphi \rangle \\ \leq \|(1 - \vartheta_R) \nabla \varrho_\varepsilon * \widetilde{w}\|_{L^2(\mathbb{R}^n)} + C \|\nabla \varrho_\varepsilon * \widetilde{w} \, \mathbf 1_{A_R}\|_{L^2(\mathbb{R}^n)} + \|\nabla \varrho_\varepsilon * \widetilde{w} - \nabla \widetilde{w}\|_{L^2(\mathbb{R}^n)} \end{eqnarray} |
and so
\begin{equation} \lim\limits_{\varepsilon\to 0^+} \lim\limits_{R \to \infty} \|w_{\varepsilon, R} - w\|_{D^{-1}(\mathbb{R}^n)} = 0 . \end{equation} | (2.5) |
Next we estimate TV(w_{\varepsilon, R}) . Due to lower semicontinuity of TV , we can assume without loss of generality that TV(w) < \infty . First, we note that \varrho_\varepsilon * w \in D^{-1}(\mathbb{R}^n) \cap C^\infty(\mathbb{R}^n) ([29,30]) and
\|\varrho_\varepsilon * \psi\|_{L^\infty(\mathbb{R}^n, \mathbb{R}^n)} \leq \|\psi\|_{L^\infty(\mathbb{R}^n, \mathbb{R}^n)}. |
Thus, for any \psi \in L^\infty(\mathbb{R}^n, \mathbb{R}^n) ,
TV(\varrho_\varepsilon * w) = \sup\limits_{\psi \in X_1} \left\{ \langle w, \operatorname{div}\, \varrho_\varepsilon *\psi\rangle \right\} \leq TV(w). |
In particular, this implies that \nabla \varrho_\varepsilon * w \in L^1(\mathbb{R}^n) for \varepsilon > 0 and \int |\nabla \varrho_\varepsilon * w| = TV(\varrho_\varepsilon * w) \leq TV(w) . By Lemma 8,
\begin{eqnarray} TV(w_{\varepsilon, R}) = \int|\nabla w_{\varepsilon, R}| \leq \int\vartheta_R\left|\nabla \varrho_\varepsilon * w \right| + \int\left|\left(\varrho_\varepsilon * w - \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R}\right) \nabla \vartheta_R\right| \\ \leq \int\left|\nabla \varrho_\varepsilon * w \right| + C \left\|\varrho_\varepsilon * w - \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R}\right\|_{L^{1^*}(\mathbb{R}^n)} \|\nabla \vartheta_R\|_{L^n(\mathbb{R}^n)} \\ \leq \|\nabla \varrho_\varepsilon * w\|_{L^1(\mathbb{R}^n)} + C \|\nabla \varrho_\varepsilon * w\|_{L^1(\mathbb{R}^n)} R^{-\frac{(n-1)(n-2)}{n}} \leq \left(1 + CR^{-\frac{(n-1)(n-2)}{n}}\right) TV(w). \end{eqnarray} | (2.6) |
If n\geq 3 , together with lower semicontinuity of TV , this yields
\lim\limits_{(\varepsilon, R) \to (0, \infty)} TV(w_{\varepsilon, R}) = TV(w). |
Taking into account (2.5), by a diagonal procedure we can select sequences (\varepsilon_k) , (R_k) such that w_k : = w_{\varepsilon_k, R_k} satisfies both requirements in the assertion. On the other hand, if n = 1 or n = 2 , (2.6) only implies uniform boundedness of TV(\nabla w_{\varepsilon, R}) .
Let now n = 2 . Since w_{\varepsilon, R} have compact support, uniform boundedness of TV(\nabla w_{\varepsilon, R}) implies uniform bound on w_{\varepsilon, R} in L^2(\mathbb{R}^n) by the Sobolev inequality. As w_{\varepsilon, R} converges to \varrho_\varepsilon * w in D^{-1}(\mathbb{R}^n) , we have \varrho_\varepsilon * w \in L^2(\mathbb{R}^n) . This allows us to improve (2.6):
\begin{eqnarray} TV(w_{\varepsilon, R}) = \int|\nabla w_{\varepsilon, R}| \leq \int\vartheta_R\left|\nabla \varrho_\varepsilon * w \right| + \int\left|\varrho_\varepsilon * w \nabla \vartheta_R\right| + \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R} \int|\nabla \vartheta_R| \\ \leq \|\nabla \varrho_\varepsilon * w \|_{L^1(\mathbb{R}^2)} + C\|\varrho_\varepsilon * w\, \mathbf 1_{A_R}\|_{L^2(\mathbb{R}^2)} \|\nabla \vartheta_R\|_{L^2(\mathbb{R}^2)} + C\frac{\|\vartheta_R\|_{D_0^1(\mathbb{R}^2)}\| \varrho_\varepsilon * w\|_{D^{-1}(\mathbb{R}^2)}}{\|\vartheta_R\|_{L^1(\mathbb{R}^2)}} \|\nabla\vartheta_R\|_{L^1(\mathbb{R}^2)} \\ \leq TV(w) + C\|\varrho_\varepsilon * w\, \mathbf 1_{A_R}\|_{L^2(\mathbb{R}^2)} + \frac{C}{R}\| \varrho_\varepsilon * w\|_{D^{-1}(\mathbb{R}^2)}. \end{eqnarray} | (2.7) |
The r. h. s. of (2.7) converges to TV(w) as R\to \infty and we conclude as before.
Next, consider n = 1 . In this case, finiteness of TV(w) implies that \varrho_\varepsilon * w \in L^\infty(\mathbb{R}^n) and there exist g_\varepsilon^\pm \in \mathbb{R} such that
\lim\limits_{x \to \pm \infty} \varrho_\varepsilon * w(x) = g_\varepsilon^\pm. |
Now, let \eta^\pm_R be the element of minimal norm in D_0^1(\mathbb{R}) under constraints
\eta^\pm_R(\pm x) = 1 \text{ if } x \in [2R, 3R], \quad \eta^\pm_R(\pm x) = 0 \text{ if } x \not \in ]R, 4R[ . |
(Clearly, \eta^\pm_R is a continuous, piecewise affine function.) We have
|\langle \varrho_\varepsilon * w, \eta^\pm_R\rangle| \leq \|\varrho_\varepsilon * w\|_{D^{-1}(\mathbb{R})}\|\eta^\pm_R\|_{D_0^1(\mathbb{R})} \to 0 \text{ as } R \to \infty. |
On the other hand, since \eta^\pm_R are compactly supported and \varrho_\varepsilon * w coincides as distribution with a locally integrable function, we can calculate
\langle \varrho_\varepsilon * w, \eta^\pm_R\rangle = \int \varrho_\varepsilon * w\, \eta^\pm_R \to \infty \cdot g^\pm_\varepsilon \text{ as } R \to \infty, |
so g^\pm_\varepsilon = 0 . Therefore, we can estimate
\begin{eqnarray} TV(w_{\varepsilon, R}) = \int|\nabla w_{\varepsilon, R}| \leq \int\vartheta_R\left|\nabla \varrho_\varepsilon * w \right| + \int\left|\varrho_\varepsilon * w \nabla \vartheta_R\right| + \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R} \int|\nabla \vartheta_R| \\ \leq TV(w) + \frac{2}{R} \int_{A_R} |\varrho_\varepsilon * w| + 2 \frac{\int \vartheta_R \varrho_\varepsilon * w}{\int \vartheta_R}. \end{eqnarray} | (2.8) |
Since we have shown that \varrho_\varepsilon * w(x) \to 0 as x \to \pm \infty , the averages on the r. h. s. converge to 0 and we conclude as before.
As a first application of the approximation lemma, we demonstrate Lemma 6 announced before.
Proof of Lemma 6. Let w \in D(TV) and let (w_k) \subset C_{c, av}^\infty(\mathbb{R}^n) be the sequence provided by Lemma 7. Let first n > 1 . Since \nabla w_k is uniformly bounded in L^1(\mathbb{R}^n, \mathbb{R}^n) , by the Sobolev embedding w_k is uniformly bounded in L^{1^*}(\mathbb{R}^n) . Therefore, w \in L^{1^*}(\mathbb{R}^n) .
In case n = 1 , since \nabla w_k are compactly supported and uniformly bounded in L^1(\mathbb{R}) , w_k is uniformly bounded in L^\infty(\mathbb{R}) . From these two bounds it follows that w \in L^\infty(\mathbb{R}) and that \nabla w is a finite signed measure on \mathbb{R} . In particular, w has essential limits at \pm \infty . Reasoning as in the final part of the proof of Lemma 7, we show that these limits vanish.
For a gradient flow of a convex functional, there is a general theory initiated by Y. K\overline{\rm o}mura [24] and developed by H. Brézis [6] and others. It is summarized as follows.
Proposition 9 ([6]). Let H be a real Hilbert space. Let \mathcal{E} be a lower semicontinuous, convex functional on H with values in ]\!-\!\infty, \infty] . Assume that D(\mathcal{E}) is dense in H . Then, for any u_0\in H , there exists a unique solution u\in C\left([0, \infty[\, , H\right) which is absolutely continuous in (\delta, T) (for any \delta < T < \infty ) satisfying
\begin{equation} \left \{ \begin{array}{l} u_t \in -\partial\mathcal{E}(u) \quad\mathit{\text{for a.e.}}\quad t > 0 \\ u(0) = u_0. \end{array} \right. \end{equation} | (3.1) |
Moreover,
\int^t_s \| u_t \|^2_H\, d\tau \leq \mathcal{E}\left(u(s)\right) - \mathcal{E}\left(u(t)\right) \quad\mathit{\text{for all}}\quad t \geq s > 0. |
If \mathcal{E}(u_0) < \infty , then s = 0 is allowed. In particular, u_t\in L^2(0, \infty; H) .
As in [2], this solution satisfies the evolutionary variational inequality
\frac12 \frac{\mathrm{d}}{\mathrm{d}t} \|u-f\|^2_H \leq \mathcal{E}(f) - \mathcal{E}(u) \quad\text{for a.e.}\ t > 0 |
for any f\in H . Indeed, by definition, u_t\in-\partial\mathcal{E}(u) is equivalent to saying
\frac12 \frac{\mathrm{d}}{\mathrm{d}t} \|u-f\|^2_H = (-u_t, f-u)_H \leq \mathcal{E}(f) - \mathcal{E}(u) |
for any f\in H . The evolutionary variational inequality is not only an equivalent formulation of the gradient flow u_t\in-\partial\mathcal{E}(u) , but also apply to a gradient flow of a metric space by replacing \|u-f\|_H by distance between u and f ; see [2] for the theory.
To be able to actually find solutions to (3.1), we need to characterize the subdifferential of the total variation in the space D^{-1} = D^{-1}(\mathbb{R}^n) . The basic idea of the proof is a duality argument, which has been carried out in the case of L^2 subdifferentials. In the case of L^2 setting, the idea goes back to the unpublished note of F. Alter and a detailed proof is given in [3]. Let \mathcal{E} be a functional on a real Hilbert space H equipped with an inner product (\cdot, \cdot)_H . The main idea is to characterize the subdifferential \partial\mathcal{E} by the polar \mathcal{E}^0 of \mathcal{E}:H\to[-\infty, \infty] which is defined by
\begin{equation*} \mathcal{E}^0(v) : = \sup\left\{(u, v)_H \bigm| u\in H, \ \mathcal{E}(u)\leq 1 \right\} = \sup\left\{(u, v)/\mathcal{E}(u) \bigm| u\in D(\mathcal{E}), \ \mathcal{E}(u)\neq 0 \right\}, \end{equation*} |
where D(\mathcal{E}) = \left\{u\in H \bigm| |\mathcal{E}(u)| < \infty\right\} . We first recall a lemma [3, Lemma 1.7].
Lemma 10. Let \mathcal{E} be convex. Assume that \mathcal{E} is positively one-homogeneous, i.e.,
\mathcal{E}(\lambda u) = \lambda\mathcal{E}(u) |
for all \lambda > 0 , u\in H . Then, v\in\partial\mathcal{E}(u) if and only if \mathcal{E}^0(v)\leq 1 and (u, v)_H = \mathcal{E}(u) .
Remark 11. By general theory of convex functionals, we know that
(\mathcal{E}^0)^0 = \mathcal{E} |
if \mathcal{E} is a non-negative, lower semicontinuous, convex, positively one-homogeneous functional [3, Proposition 1.6].
This property is essential for the proof of
Theorem 12. Let \Psi\colon D^{-1} \to [0, \infty] by defined by
\Psi(v) = \inf \left\{ \|Z\|_\infty \bigm| v = \Delta\operatorname{div}Z, \ Z \in L^\infty(\mathbb{R}^n), \ \operatorname{div}Z \in D^1_0 \right\}. |
Then (TV)^0 = \Psi .
Remark 13. (i) By definition, \Psi is a convex, lower semi-continuous, positively one-homogeneous function, so (\Psi^0)^0 = \Psi .
(ii) if \Psi(v) < \infty , the infimum is attained. Theorem 12 together with Lemma 10 implies the following characterization of the subdifferential of TV .
Theorem 14. An element v\in D^{-1} belongs to \partial TV(u) if and only if there is Z\in L^\infty(\mathbf{R}^n) with \operatorname{div}Z\in D^1_0 such that
(i) |Z| \leq 1
(ii) v = \Delta \operatorname{div}Z
(iii) -\langle u, \operatorname{div}Z \rangle = TV(u) .
Proof. By Lemma 10 and Theorem 12,
v \in \partial TV(u)\ \iff\ \Psi(v) \leq 1 \text{ and } (v, u)_{D^{-1}} = TV(u). |
The property \Psi(v)\leq1 together with Remark 13(ii) implies (i), (ii) and \operatorname{div}Z\in D^1_0 .
(v, u)_{D^{-1}} = \langle u, (-\Delta)^{-1}v \rangle = -\langle u, \operatorname{div}Z \rangle. |
It is not difficult to check the converse.
Proof of Theorem 12. The inequality TV^0\leq\Psi :
We take v\in D^{-1} with \Psi(v) < \infty . By Remark 13(ii), there is Z\in L^\infty(\mathbb{R}^n) with v = \Delta\operatorname{div}Z with \operatorname{div}Z\in D^1_0 such that \Psi(v) = \|Z\|_\infty . By Lemma 7, there is u_k\in C^\infty_{c, av} such that TV(u_k)\to TV(u) , u_k\to u in D^{-1} . We observe that
\begin{align*} (u_k, v)_{D^{-1}} & = \langle u_k, (-\Delta)^{-1}v \rangle = -\langle u_k, \operatorname{div}Z \rangle \\ & = \int_{\mathbb{R}^n} Z \cdot \nabla u_k \leq \|Z\|_\infty TV(u_k). \end{align*} |
Sending k\to\infty , we conclude that
(u, v)_{D^{-1}} \leq \|Z\|_\infty \quad\text{for all}\quad u \in D^{-1} \quad\text{with}\quad TV(u) \leq 1. |
By definition of \Psi , this implies TV^0\leq\Psi .
The inequality \Psi\leq TV^0 :
By definition,
\begin{align*} TV(u) & = \sup \left\{ \langle u, -\operatorname{div}Z \rangle \bigm| Z \in C^\infty_c(\mathbb{R}^n), \ |Z| \leq 1 \right\} \\ & = \sup \left\{ \frac{\langle u, -\operatorname{div}Z \rangle}{\|Z\|_\infty} \biggm| Z \in C^\infty_c(\mathbb{R}^n), \ Z \neq 0 \right\}. \end{align*} |
Since
\langle u, -\operatorname{div}Z \rangle = \langle u, (-\Delta)^{-1} \Delta\operatorname{div}Z \rangle = ( u, \Delta\operatorname{div}Z )_{D^{-1}}, |
we observe that
\begin{align*} TV(u) & = \sup \left\{ \frac{( u, \Delta\operatorname{div}Z )_{D^{-1}}}{\|Z\|_\infty} \biggm| Z \in C^\infty_c(\mathbb{R}^n), \ Z \neq 0 \right\} \\ &\leq \sup \left\{ \frac{( u, \Delta\operatorname{div}Z )_{D^{-1}}}{\Psi(\Delta\operatorname{div}Z)} \biggm| Z \in C^\infty_c(\mathbb{R}^n), \ \Psi(\Delta\operatorname{div}Z) \neq 0 \right\} \\ &\leq \Psi^0(u). \end{align*} |
This implies that TV^0\geq(\Psi^0)^0 = \Psi .
Now that the subdifferential of TV in D^{-1} is calculated, we are able to justify an explicit definition of a solution proposed in [15].
Theorem 15. Assume that u\in C\left([0, \infty[\, , D^{-1}\right) . Then u is a solution of u_t\in-\partial_{D^{-1}}TV(u) with u_0 = u(0) in the sense of Proposition 9 if and only if there exists Z\in L^\infty\bigl(]0, \infty[\times\mathbb{R}^n\bigr) satisfying
\operatorname{div}Z \in L^2 \left(\delta, \infty; D^1_0 (\mathbb{R}^n)\right) \quad\mathit{\text{for any}}\quad \delta > 0 |
such that for a.e. t\in]0, \infty[ there holds
u_t = -\Delta\operatorname{div}Z \quad\mathit{\text{in}}\quad D^{-1} (\mathbb{R}^n), |
|Z| \leq 1 \quad \mathcal{L}^n\mathit{\text{-a.e.}} |
and
\langle u, \operatorname{div}Z \rangle = -TV(u). |
(If TV(u_0) < \infty , \delta = 0 is allowed.)
The Theorem essentially follows from Theorem 14. We only need to justify that a Cahn-Hoffman vector field Z defined separately for every time instance by Theorem 14 can be chosen to be jointly measurable, i.e., Z\in L^\infty\bigl(]0, \infty[\times\mathbb{R}^n\bigr) . As in the second-order case [3, section 2.4], this can be done by recalling that Bochner functions can be well approximated by piecewise constant functions. Since our situation is slightly different, let us include the argument for completeness. We begin with a lemma which is a version of [3, Lemma A.8] suited to our needs.
Lemma 16. Let v \in L^1_{loc}(]0, \infty[, X) , where X is a Banach space and let N \subset]0, \infty[ have Lebesgue measure 0 . Then for each \varepsilon > 0 there exists a countable family \mathcal{G} of disjoint closed intervals I_k = \overline{B(t_k, r_k)} , k \in \mathbb{N} , such that t_k is a Lebesgue point for v , t_k \not \in N ,
\mathcal{L}^1 \left(]0, \infty[\setminus \bigcup\limits_{k \in \mathbb{N}} I_k\right) = 0 |
and
\int_0^\infty \|v - v^\varepsilon \|_X \leq \varepsilon, \quad \mathit{\text{where}}\; v^\varepsilon (t) = v(t_k) \quad \mathit{\text{for}}\; t \in I_k, \ k \in \mathbb{N}. |
Proof. Referring e.g., to [6, p. 140], \mathcal{L}^1 -a.e. point t\in]0, \infty[ is a Lebesgue point for v , i.e.,
\lim\limits_{h \to 0^+} \frac{1}{2h} \int_{t-h}^{t+h} \|v - v(t)\|_X = 0. |
Let A be the set of Lebesgue points of v contained in ]0, \infty[\setminus N and let us take
\mathcal{F} = \Bigg\{\overline{B(t, r)}\ \Bigg|\ t \in A, \ r < \min(\varepsilon, t), \ \frac{1}{2r} \int_{t-r}^{t+r} \|v - v(t)\|_X \leq \varepsilon e^{-t} \Bigg\}. |
Using Besicovitch covering theorem [9, Corollary 1 on p. 35] with U = ]0, \infty[ and (for example) \mathrm{d}\mu(t) = e^{-t}\, \mathrm{d}t , we obtain a candidate for the family \mathcal{G} . We check that indeed
\int_0^\infty \|v - v^\varepsilon\|_X = \sum\limits_{k \in \mathbb{N}} \int_{I_k}\|v- v(t_k)\|_X \leq \varepsilon \sum\limits_{k \in \mathbb{N}} 2r_k \, e^{-t_k} \leq \varepsilon \int_0^\infty e^{-t} \, \mathrm{d}t = \varepsilon. |
Proof of Theorem 15. Applying Lemma 16 to u_t , for each \varepsilon > 0 we obtain a partition of ]0, \infty[ (up to a set of Lebesgue measure 0 ) into disjoint closed intervals I_k = \overline{B(t_k, r_k)} , k \in \mathbb{N} such that
\begin{equation} u_t(t_k) \in - \partial_{D^{-1}} TV(u(t_k)) \quad \text{for } k \in \mathbb{N} \end{equation} | (3.2) |
and
\int_0^\infty \|u_t - v^\varepsilon\|_{D^{-1}} \leq \varepsilon, \quad \text{where } v^\varepsilon(t) = u_t(t_k) \quad \text{for } t \in I_k, \ k \in \mathbb{N}. |
By (3.2) and Theorem 14, for k \in \mathbb{Z} there exist Z_k \in L^\infty(\mathbb{R}^n) such that
|Z_k| \leq 1, \quad u_t(t_k) = - \Delta\operatorname{div} Z_k, \quad - \langle u(t_k), \operatorname{div} Z_k \rangle = TV(u(t_k)). |
Further denoting
u^\varepsilon(t) = u(t_k), \quad Z^\varepsilon(t) = Z_k \quad \text{for } t \in I_k, \ k \in \mathbb{N}, |
we have
|Z^\varepsilon| \leq 1, \quad v^\varepsilon = - \Delta\operatorname{div} Z^\varepsilon, \quad - \langle v^\varepsilon, \operatorname{div} Z^\varepsilon \rangle = TV(u^\varepsilon) \quad \text{a.e. in } [0, \infty[. |
We immediately deduce that there exists Z \in L^\infty([0, \infty[\times \mathbb{R}^n) with |Z|\leq 1 a.e. and a subsequence Z^{\varepsilon_j} such that Z^{\varepsilon_j} converges to Z weakly-* in Z \in L^\infty([0, \infty[\times \mathbb{R}^n) . Moreover, for any \varphi \in C^\infty_c([0, \infty[\times \mathbb{R}^n) ,
\int_0^\infty \langle v^{\varepsilon_j}, \varphi\rangle = \int_0^\infty \int_{\mathbb{R}^n} \nabla \operatorname{div} Z^{\varepsilon_j} \cdot \nabla \varphi = \int_0^\infty \int_{\mathbb{R}^n} Z^{\varepsilon_j} \cdot \nabla \Delta \varphi \to \int_0^\infty \int_{\mathbb{R}^n} Z \cdot \nabla \Delta \varphi, |
at least along a subsequence. Since on the other hand
\int_0^\infty \langle v^\varepsilon, \varphi\rangle \to \int_0^\infty \langle u_t, \varphi\rangle, |
we infer that u_t = - \Delta \operatorname{div} Z and in particular \operatorname{div} Z^{\varepsilon_j} \to \operatorname{div} Z in L^1_{loc}(]0, \infty[, D^1_0) . Finally, we observe that u^\varepsilon \to u in L^\infty_{loc}(]0, \infty[, D^{-1}) . Moreover, since t \to TV(u(t)) is non-increasing, we have TV(u^\varepsilon) \to TV(u) in L^1_{loc}(]0, \infty[) . Therefore, for any \varphi \in C_c(]0, \infty[) ,
\int_0^\infty \langle u^{\varepsilon_j}, \operatorname{div} Z^{\varepsilon_j}\rangle \varphi \to \int_0^\infty \langle u, \operatorname{div} Z \rangle \varphi, \quad \int_0^\infty TV(u^\varepsilon) \varphi \to \int_0^\infty TV(u) \varphi. |
Unfortunately, in the case n\leq2 , the characteristic function \mathbf{1}_A of a set A of positive measure is not in D^{-1} since \int \mathbf{1}_A \neq0 . We shall define a new space containing \mathbf{1}_A as follows. We take a function \psi\in L^2(\mathbb{R}^n) with compact support such that \int_{\mathbb{R}^n}\psi = 1 . We introduce a vector space
E^{-1}_\psi = \left\{ w+c\psi \bigm| w \in D^{-1}(\mathbb{R}^n), \ c \in \mathbb{R} \right\}. |
This space is independent of the choice of \psi . Indeed, let \psi_i \in L^2(\mathbb{R}^n) be compactly supported and \int \psi_i = 1 ( i = 1, 2 ). An element w+c\psi_1\in E^{-1}_{\psi_1} can be rewritten as
w + c\psi_1 = w + c(\psi_1-\psi_2) + c\psi_2. |
The next lemma implies q = c(\psi_1-\psi_2)\in D^{-1}(\mathbb{R}^n) since \int q = 0 . We then conclude that w+c\psi_1\in E^{-1}_{\psi_2} .
Lemma 17. Assume that n\leq2 . A compactly supported function q\in L^2(\mathbb{R}^n) belongs to D^{-1}(\mathbb{R}^n) if and only if \int_{\mathbb{R}^n}q = 0 .
Proof. If q\in D^{-1} \cap L^1 , then
\int q = \langle q, [1] \rangle = 0, |
where [1] stands for the element of D_0^1 whose representatives are 1 as well as 0 .
Now suppose that a compactly supported function q\in L^2(\mathbb{R}^n) satisfies \int_{\mathbb{R}^n}q = 0 . Given a [\varphi] \in D_0^1 , by the Poincaré inequality, we have for any R > 0 , independently of the representative \varphi \in D^1 ,
\left\|\varphi - \frac{1}{|B_R|} \int_{B_R} \varphi\right\|_{L^2(B_R)} \leq C_R \|\nabla \varphi \|_{L^2(B_R)} \ \leq C_R \|[\varphi]\|_{D_0^1}. |
In particular, \varphi \in L^2_{loc} . Taking into account this and the assumption \int q = 0 , we see that the linear functional
\begin{equation} \langle q, [\varphi] \rangle = \int q\, \varphi \end{equation} | (4.1) |
on D_0^1 is well defined. Moreover, if R > 0 is large enough that \mathrm{supp}\, q \subset B_R ,
\left|\int q\, \varphi\right| = \left|\int_{B_R} q\left(\varphi - \frac{1}{|B_R|} \int_{B_R} \varphi\right)\right|\leq \|q\|_{L^2(B_R)} \left\|\varphi - \frac{1}{|B_R|} \int_{B_R} \varphi\right\|_{L^2(B_R)} \leq C_R \|q\|_{L^2} \|[\varphi] \|_{D_0^1}. |
Thus, the functional defined by (4.1) is bounded, i.e., q \in D^{-1} .
Since E^{-1}_\psi is independent of the choice of \psi , we suppress \psi and denote this space by E^{-1} . In case n \geq 3 , we will use notation E^{-1} = D^{-1} . We also denote E^1_0 = D^1 if n \leq 2 , E^1_0 = D^1_0 if n\geq 3 . For u \in E^{-1} , v \in E^1_0 , we denote
\begin{equation} \langle u, v \rangle_E = \left\{\begin{array}{l} \langle w, v\rangle \text{ if } n \geq 3, \\ \langle w, [v]\rangle + c \int \psi v \text{ if } n \leq 2, \end{array}\right. \end{equation} | (4.2) |
where u = w + c \psi , w \in D^{-1} , \psi \in L^2_c , \int \psi = 1 . As before, we check that the value of \langle u, v \rangle_E does not depend on the choice of this decomposition.
We recall that if n \geq 3 , E_0^1 = D_0^1 and E^{-1} = D^{-1} come with a Hilbert space structure. We also define inner products on E_0^1 , E^{-1} in case n \leq 2 by
(v_1, v_2)_{E^1_0} : = ([v_1], [v_2])_{D^1_0} + \int \psi v_1 \int \psi v_2, |
(u_1, u_2)_{E^{-1}} : = (w_1, w_2)_{D^{-1}} + c_1c_2 |
for u_i = w_i+c_i\psi , w_i\in D^{-1} , c_i\in\mathbb{R} ( i = 1, 2 ). This gives an orthogonal decomposition
E^{-1} = D^{-1} \oplus \mathbb{R}. |
We note that although the values of those products may depend on the choice of \psi , the topologies they induce on E^1_0 , E^{-1} do not. Formula (4.2) associates to any u \in E^{-1} a continuous linear functional on E^1_0 . The resulting mapping is an isometric isomorphism between E^{-1} and the continuous dual to E^1_0 .
We extend TV onto E^{-1} by defining
TV(u) : = \sup\limits_{\psi \in X_1} \langle u, -\operatorname{div}\psi \rangle_E . |
As usual, we check that TV is a convex, weakly-* (and strongly) lower semicontinuous functional. In particular, for a fixed g\in E^{-1} , the functional w\mapsto TV(w+g) is convex and lower semicontinuous on D^{-1} . We next give a definition of a solution of u_t\in-\partial TV(u) in the space E^{-1} . It turns out the idea of evolutionary variational inequality is very convenient since it is a flow in an affine space g+D^{-1} for some g\in E^{-1} .
Definition 18. Assume that u_0 \in E^{-1} . We say that u:[0, \infty[\, \to E^{-1} is a solution to
\begin{equation} u_t \in -\partial_{D^{-1}} TV(u) \end{equation} | (4.3) |
in the sense of EVI (evolutionary variational inequality) with initial datum u_0 if
(i) u-g is absolutely continuous on [\delta, T] (for any 0 < \delta < T < \infty ) with values in D^{-1} and continuous up to zero with u(0) = u_0 and
(ii) u-g satisfies the evolutionary variational inequality, i.e.,
\frac12 \frac{\mathrm{d}}{\mathrm{d}t} \left\| u(t)-g \right\|^2_{D^{-1}} \leq TV(g) - TV \left(u(t)\right) \quad |
holds for a.e. t > 0 , provided that g\in E^{-1} is such that u_0-g\in D^{-1} .
Theorem 19. For any u_0\in E^{-1} , there exists a unique solution u of (4.3) in the sense of EVI. Moreover, if u^i is the solution to (4.3) in the sense of EVI with u^i(0) = u^i_0 \in E^{-1} for i = 1, 2 , then
\begin{equation} \left\| u^1(t) - u^2(t) \right\|_{D^{-1}} \leq \left\| u^1_0 - u^2_0\right\|_{D^{-1}} \quad\mathit{\text{for all}}\quad t\geq 0, \end{equation} | (4.4) |
provided that u^1_0 - u^2_0 \in D^{-1} .
Proof. Uniqueness follows from contractivity (4.4), which is established by a standard reasoning [2]. We give a short proof for the reader's convenience and for completeness. Let u^i ( i = 1, 2 ) be a solution to (4.3) in the sense of EVI with initial datum u^i_0 such that u^1_0 - u^2_0\in D^{-1} . Since u^i are solutions, we also have u^i(t)-u^i_0\in D^{-1} for t\geq0 and so u^1_0 - u^2(t)\in D^{-1} , u^2_0 - u^1(t)\in D^{-1} . Thus, EVI yields
\begin{eqnarray} \frac12\frac{\mathrm{d}}{\mathrm{d}t} \left\| u^1-u^2 \right\|^2_{D^{-1}} = \left(u^1_t, u^1 - u^2\right)_{D^{-1}} + \left(u^2_t, u^2 - u^1\right)_{D^{-1}} \\ = \left.\frac12\frac{\mathrm{d}}{\mathrm{d}t} \left\| u^1(t)-g \right\|^2_{D^{-1}}\right|_{g = u^2(t)} + \left.\frac12\frac{\mathrm{d}}{\mathrm{d}t} \left\| u^2(t)-g \right\|^2_{D^{-1}}\right|_{g = u^1(t)} \\ \leq TV\left(u^2(t)\right)-TV\left(u^1(t)\right) + TV\left(u^1(t)\right)-TV\left(u^2(t)\right) = 0 \end{eqnarray} |
for a.e. t > 0 . We conclude that \left\| u^1(t)-u^2(t) \right\|^2 is non-increasing, in particular (4.4) holds.
The existence is a bit more involved. For u_0 = w_0+g_0\in E^{-1} with w_0\in D^{-1} , we consider the gradient flow of the form
\begin{equation} w_t \in -\partial_{D^{-1}} TV(w+g_0), \quad w(0) = w_0. \end{equation} | (4.5) |
Applying Proposition 9, there is a unique solution w to (4.5) for w_0\in D^{-1} . This solution satisfies the evolutionary variational inequality
\frac12\frac{\mathrm{d}}{\mathrm{d}t} \| w-f \|^2_{D^{-1}} \leq TV(f+g_0) - TV(w+g_0) \quad\text{for a.e. } t > 0 |
for any f\in D^{-1} . Setting u = w+g_0 , g = f+g_0 , we end up with
\frac12\frac{\mathrm{d}}{\mathrm{d}t} \| w-g \|^2_{D^{-1}} \leq TV(g) - TV\left(u(t)\right). |
Since g can be taken arbitrary such that u_0-g\in D^{-1} , this shows that u is the solution of (4.3) in the sense of EVI; condition (i) follows easily from Proposition 9.
It is non-trivial to characterize the subdifferential \partial_{D^{-1}}TV . For this purpose, we introduce a mapping I which plays a role analogous to -\Delta in n \geq 3 .
Lemma 20. Let n \leq 2 . The mapping I\colon E^1_0 \to E^{-1} defined by
I(f) = (-\Delta) [f] + \left( \int_{\mathbb{R}^n} f\psi \right) \psi |
is an isometric isomorphism.
Proof. It is clear that I(f)\in E^{-1} and I is linear. For given u = w+c\psi\in E^{-1} with w\in D^{-1} , c\in\mathbb{R} , there is \overline{f}\in D^1_0 such that (-\Delta)\overline{f} = w . Since a representative f of \overline{f} is determined up to an additive constant, there is a unique representative f such that
\int_{\mathbb{R}^n} f\psi = c. |
Thus, the mapping I is surjective. If I(f) = 0 , then (-\Delta) [f] = 0 so [f] = 0 . Thus f is a constant. Since \int f\, \psi = 0 , this constant must be zero, so f = 0 . Thus, I is injective. Recalling our definitions of inner products on E^1_0 , E^{-1} , it is easy to check that I is an isometry.
We have a characterization of the polar of TV in E^{-1} as in Theorem 12.
Theorem 21. Let n \leq 2 . Let \Psi be given by
\Psi(v) = \inf \left\{ \|Z\|_\infty \bigm| v = I(-\operatorname{div} Z), \ Z \in L^\infty(\mathbb{R}^n), \ \operatorname{div}Z \in E_0^1 \right\} |
for v\in E^{-1} . Then (TV)^0 = \Psi .
Admitting this fact, we are able to give a characterization of the subdifferential.
Theorem 22. Let n \leq2 . An element v \in E^{-1} belongs to \partial_{E^{-1}}TV(u) if and only if there is Z\in L^\infty(\mathbb{R}^n) with \operatorname{div}Z\in E_0^1 such that
(i) |Z| \leq 1
(ii) v = I(-\operatorname{div}Z)
(iii) -\langle u, \operatorname{div}Z \rangle_E = TV(u).
Proof of Theorem 22. The proof parallels that of Theorem 14. By Lemma 10 and Theorem 21
v \in \partial TV(u)\ \iff\ \Psi(v) \leq 1 \text{ and } (u, v)_{E^{-1}} = TV(u). |
The properties (i), (ii) together with \operatorname{div}Z\in E^1_0 are equivalent to \Psi(v)\leq1 . Since
\begin{equation} (u, v)_{E^{-1}} = \left(w, (-\Delta)[v] \right)_{D^{-1}} + c \int_{\mathbb{R}^n} v\, \psi = \langle w, [v] \rangle + c \int_{\mathbb{R}^n} v\, \psi, \end{equation} | (4.6) |
the Euler equation (u, v)_{E^{-1}} = TV(u) is equivalent to (iii).
Proof of Theorem 21. The proof parallels that of Theorem 12. We first prove that
(u, v)_{E^{-1}} \leq \|Z\|_\infty \quad\text{for all}\quad u \in E^{-1} \quad\text{with}\quad TV(u) \leq 1 |
for v = I(-\operatorname{div}Z) . This implies TV^0\leq\Psi . The estimate (u, v)_{E^{-1}}\leq \|Z\|_\infty formally follows from the identity (4.6). Indeed, by (4.6), we see
(u, v)_{E^{-1}} = - \langle w, [\operatorname{div}Z]\rangle - c \int_{\mathbb{R}^n} \psi \operatorname{div}Z. |
If u is in C^\infty_c({\mathbb{R}^n}) , then, by this formula, we obtain
(u, v)_{E^{-1}} = -\int_{\mathbb{R}^n} u\operatorname{div}Z = \int_{\mathbb{R}^n} \nabla u \cdot Z \leq \|Z\|_\infty TV(u). |
By approximation, as in the proof of Theorem 12, we conclude the desired estimate.
The other inequality \Psi\leq TV^0 follows from TV\leq\Psi^0 . The proof of TV\leq\Psi^0 is parallel to that of Theorem 12 by replacing \Delta\operatorname{div}Z by I(-\operatorname{div}Z) and the D^{-1} inner product by the E^{-1} inner product, respectively, if one notes the identity (4.6). Since I is an isometry, \Psi is lower semicontinuous, and we conclude that \Psi = TV^0 by Remark 11.
We have to be careful, since the E^{-1} gradient flow
u_t \in -\partial_{E^{-1}} TV(u) |
does not correspond to the total variation flow u_t = (-\Delta)\operatorname{div}\left(\nabla u/|\nabla u|\right) . By Theorem 22(iii), Z = \nabla u/|\nabla u| if \nabla u\neq0 . Thus the E^{-1} gradient flow is formally of the form
u_t = (-\Delta)\operatorname{div}\left(\nabla u/|\nabla u|\right) + \psi \int_{\mathbb{R}^n} \psi \operatorname{div}\left(\nabla u/|\nabla u|\right). |
To recover the original total variation flow, we consider "partial" subdifferential in the direction of D^{-1} . Let P be the orthogonal projection from E^{-1} to D^{-1} . Then, by definition,
\partial_{D^{-1}} TV(w+c\psi) = P \partial_{E^{-1}} TV(u). |
The equation
w_t \in -\partial_{D^{-1}} TV(w+c\psi) |
is now formally of the form
u_t = (-\Delta) \operatorname{div}\left(\nabla u/|\nabla u|\right) |
since c\psi is time-independent. Here is a precise statement.
Theorem 23. Let n \leq 2 . Consider the functional \mathcal{F}:w\mapsto TV(w+c\psi) on D^{-1} for a fixed c\in\mathbb{R} and \psi . Then, \partial_{D^{-1}}\mathcal{F}(w) = P\partial_{E^{-1}}TV(u) for u = w+c\psi . In particular, an element v\in D^{-1} belongs to \partial_{D^{-1}}\mathcal{F}(w) if and only if there is Z\in L^\infty(\mathbb{R}^n) with \operatorname{div}Z\in E^1_0 such that
(i) |Z| \leq 1 ,
(ii) v = \Delta\operatorname{div}Z ,
(iii) - \langle u, \operatorname{div}Z \rangle_E = TV(u) .
(In case n\leq 2 , by \Delta\operatorname{div}Z we understand \Delta[\operatorname{div}Z] .) This characterization is important to calculate the solution of u_t = (-\Delta)\operatorname{div}\left(\nabla u/|\nabla u|\right) for n\leq2 explicitly. In fact, we can recover the characterization of a solution in the sense of EVI analogous to the one in Theorem 15, amounting to Theorem 2.
Proof of Theorem 2. This is almost immediate. However, like in the case of Theorem 15, we need to justify that the vector field Z can be chosen to be jointly measurable with respect to (t, x) . We proceed as in the proof of Theorem 15. The difference is, now we need to pass to the limit with
- \langle u^\varepsilon, \operatorname{div} Z^\varepsilon \rangle_E = TV(u^\varepsilon), |
but convergence of v^\varepsilon in L^1_{loc}(]0, \infty[, D^{-1}) gives only [\operatorname{div} Z^\varepsilon] \to [\operatorname{div} Z] \in L^1_{loc}(]0, \infty[, D^1_0) . To deal with this problem, let us choose \psi \in C^1_c(\mathbb{R}^n) and let c \in \mathbb{R} be such that u(t) - c \psi \in D^{-1} for t > 0 . Then we also have u^\varepsilon(t) - c \psi \in D^{-1} for t > 0 , \varepsilon > 0 and
\langle u^\varepsilon, \operatorname{div} Z^\varepsilon \rangle_E = \langle u^\varepsilon - c \psi, [\operatorname{div} Z^\varepsilon] \rangle + c \int_{\mathbb{R}^n} \psi \operatorname{div} Z^\varepsilon = \langle u^\varepsilon - c \psi, [\operatorname{div} Z^\varepsilon] \rangle - c \int_{\mathbb{R}^n} \nabla \psi \cdot Z^\varepsilon. |
Testing with \varphi \in C_c(]0, \infty[) and using weak-* convergence of Z^\varepsilon in L^\infty(]0, \infty[\times \mathbb{R}^n) we get
\begin{eqnarray} \int_0^\infty \langle u^\varepsilon, \operatorname{div} Z^\varepsilon \rangle_E \, \varphi = \int_0^\infty \langle u^\varepsilon - c \psi, [\operatorname{div} Z^\varepsilon] \rangle \varphi - c \int_0^\infty \int_{\mathbb{R}^n} \nabla \psi \cdot Z^\varepsilon \varphi \\ \to \int_0^\infty \langle u - c \psi, [\operatorname{div} Z] \rangle \varphi - c \int_0^\infty \int_{\mathbb{R}^n} \nabla \psi \cdot Z \varphi = \int_0^\infty \langle u, \operatorname{div} Z \rangle_E \, \varphi, \end{eqnarray} |
at which point we conclude as in the proof of Theorem 15.
We are interested in sets where the speed of solution u_t is spatially constant. The speed is given as minus the minimal section of the subdifferential, i.e.,
\partial^0_{D^{-1}} TV(u) : = \arg\min \left\{ \|v\|_{D^{-1}} \bigm| v \in \partial_{D^{-1}} TV(u) \right\}. |
Since \partial_{D^{-1}} TV(u) is closed and convex, \partial^0_{D^{-1}} TV(u) is uniquely determined if \partial_{D^{-1}} TV(u)\neq\emptyset . Since we have characterized the subdifferential, we end up with
\begin{eqnarray} \partial^0_{D^{-1}} TV(u) = \arg\min \bigl\{ \|v\|_{D^{-1}} \bigm| v = \Delta \operatorname{div}Z, \ Z \in L^\infty(\mathbb{R}^n, \mathbb{R}^n), \ |Z| \leq 1, \\ \operatorname{div}Z \in E^1_0(\mathbb{R}^n), \ -\langle u, \operatorname{div}Z \rangle_E = TV(u)\bigr\}. \end{eqnarray} |
Although the minimizer v is unique, the corresponding Z may not be unique. Let U be a smooth open set in \mathbb{R}^n . We consider a smooth function u such that
\overline{U} = \left\{ x \in \mathbb{R}^n \bigm| u(x) = 0 \right\} |
and \partial_{D^{-1}} TV(u)\neq\emptyset . Such a closed set is often called a facet. Assume further that \nabla u\neq0 outside \overline{U} . Let Z be a vector field satisfying v = \Delta\operatorname{div}Z for v\in\partial TV(u) . It is easy to see that outside the facet \overline{U} ,
Z(x) = \nabla u(x) \bigm/ \left| \nabla u(x) \right| |
by - \langle u, \operatorname{div}Z \rangle_E = TV(u) . Since \|v\|_{D^{-1}} = \|\operatorname{div}Z\|_{D^1_0} , we see that
\partial^0_{D^{-1}} TV(u) = \arg\min \left\{ \|\operatorname{div}Z\|_{D^1_0} \bigm| |Z| \leq 1\ \text{in}\ U, \ Z = \nabla u / |\nabla u|\ \text{in}\ \overline{U}^c, \ \operatorname{div}Z \in E^1_0 \right\}. |
Since \operatorname{div}Z is locally integrable, the normal trace is well-defined from inside as an element of L^\infty(\partial U) [3] and it must agree with that from outside, i.e.,
\begin{equation} \nu \cdot Z = \nu \cdot \nabla u \bigm/ |\nabla u| = \nu \cdot \chi\nu = \chi, \end{equation} | (5.1) |
where \nu(x) is the exterior unit normal of \partial U and
\begin{equation*} \chi(x) = \left \{ \begin{array}{cl} 1 &\text{if}\ u > 0\ \text{outside}\ \overline{U}\ \text{near}\ x \in \partial U, \\ -1 &\text{otherwise.} \end{array} \right. \end{equation*} |
Since \operatorname{div}Z is in E^1_0 , its trace from inside and outside must agree, i.e.,
\operatorname{div}Z = \operatorname{div} \left(\nabla u/|\nabla u|\right) = \chi \operatorname{div}\nu = : \chi \kappa \quad \text{on }\partial U. |
Note that \kappa(x) is the sum of all principal curvatures, equal to n-1 times the (inward) mean curvature, of \partial U at x .
Let Z_0 be a minimizer corresponding to v = \partial^0_{D^{-1}} TV(u) . Since the value Z_0 outside \overline{U} is always the same, we consider its restriction on U and still denote by Z_0 . Then, Z_0 is a minimizer of
\begin{equation} \left\{ \int_U |\nabla\operatorname{div}Z|^2 \bigm| |Z| \leq 1\ \text{in}\ U, \ \nu\cdot Z = \chi\ \text{on}\ \partial U, \ \operatorname{div}Z = \chi \kappa\ \text{on}\ \partial U \right\}. \end{equation} | (5.2) |
Although \operatorname{div}Z_0 \in D^1_0(\mathbb{R}^n) so that \nabla\operatorname{div}Z_0 \in L^2(\mathbb{R}^n, \mathbb{R}^n) , the quantity \nabla\operatorname{div}Z_0 may jump across \partial U . Thus \Delta\operatorname{div}Z_0 may contain singular part which is a driving force to move the facet boundary "horizontally" during its evolution under the fourth-order total variation equation as observed in the previous section and earlier in [12]. In the second-order problem, the speed does not contain any singular part so the jump discontinuity does not move.
We are interested in a situation where \Delta\operatorname{div}Z_0 is constant over U . In the spirit of [25], we call any continuous function \chi \colon \partial U \to \{-1, 1\} a signature for U .
Definition 24. Let U be a smooth open set in \mathbb{R}^n with signature \chi . We say that U is ( D^{-1} -)calibrable (with signature \chi ) if there exists Z_0 satisfying the constraint
\begin{equation} |Z_0|\leq 1 \quad {on}\quad U \end{equation} | (5.3) |
with boundary conditions
\begin{equation} \nu\cdot Z_0 = \chi, \quad \operatorname{div}Z_0 = \chi \kappa\ \quad{on}\quad \partial U, \end{equation} | (5.4) |
with the property that
\begin{equation} \Delta\operatorname{div}Z_0 \;{ is\; constant\; over }\; U. \end{equation} | (5.5) |
We call any such Z_0 a ( D^{-1} -)calibration for U (with signature \chi ).
From the definition of calibration, we easily deduce
Proposition 25. Let U be a smooth bounded domain in \mathbb{R}^n . Assume that Z_0 is a calibration for U with signature \chi . Then, w_0 = \operatorname{div}Z_0 is a solution to the Saint-Venant problem
\left\{\begin{array}{l} -\Delta w = \lambda \; { in }\; U\quad\quad\quad\quad\quad\quad(5.6)\\ w = \chi^K \; { on }\; \partial U\quad\quad\quad\quad\quad\quad(5.7) \end{array}\right. |
with the constraint
\begin{equation} \int_U w \, \mathrm{d} \mathcal{L}^n = \int_{\partial U}\chi \, \mathrm{d} \mathcal{H}^{n-1}, \end{equation} | (5.8) |
where \lambda is some constant.
Appealing to this relation between calibrability and the Saint-Venant problem, we can prove the following
Theorem 26. Let U be a smooth bounded domain in \mathbb{R}^n . Suppose that Z^* is a calibration for U with signature \chi . Then Z^* is a minimizer of (5.2).
Proof. We first note that w_* = \operatorname{div}Z_* must satisfy (5.8). We consider the minimization problem for
e(w) = \int_U |\nabla w|^2 |
under the Dirichlet condition (5.7) and the constraint (5.8). Since the problem is strictly convex, there is a unique minimizer \overline{w} in D^1_0(U) . By Lagrange's multiplier method, \overline{w} must satisfy (5.6) because of the constraint (5.8). (Actually, a weak solution \overline{w} of (5.6) is a smooth solution of (5.6), (5.7) by the standard regularity theory of linear elliptic partial differential equations [19, Chapter 6].) As we see in Lemma 27 below, the constant \lambda is uniquely determined by (5.6) and (5.8). For \overline{w} , there always exists Z\in C^\infty(\overline{U}) such that
\begin{equation} \operatorname{div}Z = \overline{w} \quad\text{in}\quad U, \quad \nu \cdot Z = \chi \quad\text{on}\quad \partial U. \end{equation} | (5.9) |
Indeed, let p be a solution of the Neumann problem
\Delta p = \overline{w} \quad\text{in}\quad U, \quad \nu \cdot \nabla p = \chi \quad\text{on}\quad \partial U. |
Such a solution p always exists since \overline{w} satisfies the compatibility condition (5.8) and it is smooth up to \overline{U} ; see e.g., [11,19]. If we set Z = \nabla p , then Z satisfies the desired property (5.9). Thus, the minimum e(\overline{w}) of the Dirichlet energy under the constraint (5.8) agrees with
\min \left\{ \int_U |\nabla\operatorname{div}Z|^2 \biggm| \nu \cdot Z = \chi, \ \operatorname{div}Z = \chi \kappa\ \text{on}\ \partial U \right\}. |
Since w_* = \overline{w} and |Z_*|\leq1 , this shows that Z_* is a minimizer of (5.2).
Lemma 27. Let U be a smooth bounded domain in \mathbb{R}^n . Let w solve
\begin{align*} -\Delta w & = \lambda \quad{in}\quad U \\ w & = f \quad{on}\quad \partial U \end{align*} |
for \lambda\in\mathbb{R} , f\in C(\partial U) . This solution can be written as
w = \lambda w_{\mathrm{sv}} + h_f |
where w_{\mathrm{sv}} solves the Saint-Venant problem
\begin{align} -\Delta w_{\mathrm{sv}} & = 1 \quad{in}\quad U \end{align} | (5.10) |
\begin{align} w_{\mathrm{sv}} & = 0 \quad{on}\quad \partial U \end{align} | (5.11) |
and h_f is the harmonic extension of f to U . In particular, if \int_U w = c is given then \lambda is uniquely determined by
\begin{equation} \lambda \int_U w_{\mathrm{sv}} + \int_U h_f = c, \end{equation} | (5.12) |
since w_{\mathrm{sv}} > 0 in U .
Proof. The decomposition w = \lambda w_{\mathrm{sv}}+h_f is rather clear. The property w_{\mathrm{sv}} > 0 in U follows from the maximum principle [19].
From Lemma 27 we also deduce the following formula for vertical speed of calibrable facets in terms of the solution to the Saint-Venant problem.
Proposition 28. Let U be a smooth bounded domain in \mathbb{R}^n and \lambda \in \mathbb{R} . If Z is a calibration for U with signature \chi satisfying
-\Delta \operatorname{div} Z = \lambda, |
then
\lambda \int_U w_{\mathrm{sv}} = \int_{\partial U} \chi\kappa\, \nu \cdot \nabla w_{\mathrm{sv}} + \int_{\partial U} \chi, |
where w_{\mathrm{sv}} is the solution to the Saint-Venant problem (5.10).
Proof. We recall (5.12) and calculate
\int_U h_f = - \int_U \Delta w_{\mathrm{sv}} h_f = - \int_{\partial U} \nu \cdot \nabla w_{\mathrm{sv}}\, \chi \operatorname{div} \nu + \int_U \nabla w_{\mathrm{sv}} \cdot \nabla h_f, |
\int_U \nabla w_{\mathrm{sv}} \cdot \nabla h_f = \int_{\partial U} w_{\mathrm{sv}} \nu \cdot \nabla h_f - \int_U w_{\mathrm{sv}} \Delta \nabla h_f = 0, |
c = \int_U \operatorname{div} Z = \int_{\partial U} \chi . |
We now compare the definition of calibrability for the second-order problem.
Definition 29. Let U be a smooth open set in \mathbb{R}^n with signature \chi . We say that \overline{U} is ( L^2 -)calibrable if there is Z_0 satisfying the constraint |Z_0|\leq1 in U and the boundary condition \nu\cdot Z_0 = \chi with the property that \operatorname{div}Z_0 is a constant over U .
This definition is slightly weaker than the calibrability used in [26], where a(t)\mathbf{1}_U is a solution of the total variation flow in \mathbb{R}^n with some function a(t) of t ; see also [3]. This requires that \partial^0_{L^2}TV(u) is constant not only on U but also U^c . Our definition follows from that of [5].
Definition 30. We say that a Lebesgue measurable subset U (defined up to a set of measure zero) of \mathbb{R}^n is a generalized annulus if U is non-empty, open, connected and rotationally symmetric, i.e., invariant under the linear action of SO(n) on \mathbb{R}^n .
It is easy to see that any generalized annulus is a ball, an annulus, the complement of a ball or the whole space \mathbb{R}^n . In other words, any generalized annulus is of form
A_{R_0}^{R_1} = \{x \in \mathbb{R}^n \colon R_0 < |x| < R_1\} \text{ with } 0 < R_0 < R_1 \leq \infty \quad \text{or}\quad A_0^R = B_R \text{ with }R > 0. |
In this section we will settle the question which generalized annuli are calibrable.
Lemma 31. Let U be a generalized annulus. Suppose that U is calibrable with signature \chi . Then there exists a calibration \overline{Z} for (U, \chi) of form \overline{Z}(x) = z(|x|)\frac{x}{|x|} .
Proof. Let Z be any calibration for (U, \chi) . Let \mu_n be the Haar measure on S\!O(n) . We define \overline{Z} as the average
\overline{Z}(x) = \int L Z(L^{-1} x) \, \mathrm{d} \mu_n(L). |
It is an exercise in vector calculus to check that \overline{Z} satisfies boundary conditions (5.4) and that \Delta \operatorname{div} \overline{Z} is a constant (equal to \Delta \operatorname{div} Z ) on U . By convexity, |\overline{Z}|\leq 1 . Thus \overline{Z} is a calibration for (U, \chi) . By definition, it is invariant under rotations, i.e.,
L\overline{Z}(L^{-1}x) = \overline{Z}(x) |
for L \in S\!O(n) , x \in \mathbb{R}^n . In the case n = 1 this already shows that \overline{Z} is in the desired form. In higher dimensions, we consider the orthogonal decomposition
\overline{Z}(x) = \overline{Z}\vphantom{Z}^\perp(x) + \overline{Z}\vphantom{Z}^T(x) : = \frac{x}{|x|} \otimes \frac{x}{|x|} \, \overline{Z}(x) + \left(I - \frac{x}{|x|} \otimes \frac{x}{|x|}\right) \overline{Z}(x). |
Both \overline{Z}\vphantom{Z}^\perp and \overline{Z}\vphantom{Z}^T are invariant under rotations. In particular, for any given R > 0 , the restriction of \overline{Z}\vphantom{Z}^T to \mathbb{S}^{n-1}_R is an invariant tangent vector field on \mathbb{S}^{n-1}_R . Note that any such vector field is smooth. If n = 3 , it follows by the hedgehog uncombability theorem [10, Proposition 7.15] that \overline{Z}\vphantom{Z}^T \equiv 0 . If n > 3 , any vector field invariant on \mathbb{S}^{n-1} is in particular invariant on a sphere \mathbb{S}^2 containing any given point in \mathbb{S}^{n-1} , so the same conclusion follows. Thus, we have
\overline{Z}(x) = \overline{Z}\vphantom{Z}^\perp(x) = \frac{x}{|x|} \otimes \frac{x}{|x|} \, \overline{Z}(x) = \frac{x}{|x|} \cdot \overline{Z}(x)\, \frac{x}{|x|} = : \overline{z}(x) \frac{x}{|x|}. |
By rotational invariance, we have \overline{z}(x) = z(|x|) , which concludes the proof.
We are left with the case n = 2 in which there exists a one-dimensional space of invariant tangent fields on \mathbb{S}^{n-1} = \mathbb{S}^1 spanned by e^T(x) : = (x_2, - x_1) . Thus, we have
\overline{Z}\vphantom{Z}^T(x) = z^T(|x|) e^T(x). |
We calculate
\operatorname{div} \overline{Z}\vphantom{Z}^T(x) = z^T(|x|) \operatorname{div} e^T(x) + (z^T)'(|x|) \frac{x}{|x|} \cdot e^T(x) = 0. |
Thus, we can disregard \overline{Z}\vphantom{Z}^T and choose \overline{Z}\vphantom{Z}^\perp as our calibration, since it satisfies conditions (5.3)–(5.5) (recall that \overline{Z}\vphantom{Z}^T and \overline{Z}\vphantom{Z}^\perp are orthogonal). As before, we see that
\overline{Z}\vphantom{Z}^\perp(x) = z(|x|) \frac{x}{|x|}. |
Let U be a generalized annulus. By Lemma 31, if U is calibrable, then there exists a calibration Z for U of form Z = z(|x|)\frac{x}{|x|} . It follows from (5.10) that z needs to satisfy the ODE
\begin{equation} - r^{1-n} \left(r^{n-1} \left(r^{1-n} \left(r^{n-1} z\right)'\right)'\right)' = \lambda. \end{equation} | (6.1) |
The general solution to this ODE is
\begin{equation} z(r) = c_0 r^3 + c_1 r^{3-n} + c_2 r + c_3 r^{1-n} \end{equation} | (6.2) |
where c_0 = - \frac{\lambda}{2n(n+2)} if n\neq 2 and
\begin{equation} z(r) = c_0 r^3 + c_1 r \log r + c_2 r + c_3 r^{-1}. \end{equation} | (6.3) |
where c_0 = - \frac{\lambda}{16} if n = 2 . We will now try to find a calibration for U by solving a suitable boundary value problem for (6.1).
Let U = B_R(0) . To focus attention, we choose \chi = -1 on \partial U . In this case, boundary conditions (5.4) lead to
\begin{equation} z(R) = -1, \quad z'(R) = 0. \end{equation} | (6.4) |
If n\geq 2 , in order to satisfy the requirements |Z|\leq 1 and \nabla \operatorname{div} Z , we need to restrict to c_1 = c_3 = 0 in (6.2). We make the same choice also in case n = 1 , as it leads to the right result. Then, applying (6.4) in (6.2) or (6.3), we obtain a system of two affine equations for two unknowns \lambda, c_2 . We solve it obtaining
\begin{equation} z(r) = \frac{1}{2} \left(\frac{r}{R}\right)^3 -\frac{3}{2} \frac{r}{R}, \end{equation} | (6.5) |
\begin{equation} \lambda = - \frac{n(n+2)}{R^3}. \end{equation} | (6.6) |
We check that z satisfies |z|\leq 1 on [0, R] , so Z is a calibration for B_R . Thus, all balls are calibrable in any dimension.
Let U = \mathbb{R}^n \setminus B_R . For consistency with the previous case, we choose \chi = 1 on \partial U . In this case, boundary conditions (5.4) also lead to
\begin{equation} z(R) = -1, \quad z'(R) = 0. \end{equation} | (6.7) |
Let us first assume that n \geq 3 . In order to satisfy the requirement |Z|\leq 1 , we need to restrict to \lambda = c_2 = 0 in (6.2). Again, applying (6.4) in (6.2) leads to a system of two affine equations for two unknowns c_1, c_3 . We solve it obtaining
\begin{equation} z(r) = - \frac{n-1}{2} \left(\frac{r}{R}\right)^{3-n} + \frac{n-3}{2} \left(\frac{r}{R}\right)^{1-n}. \end{equation} | (6.8) |
Again, we easily check that z satisfies |z|\leq 1 on [0, R] , so Z is a calibration for B_R .
In the omitted cases n = 1, 2 , requirement |Z|\leq 1 implies \lambda = c_1 = c_2 = 0 in (6.2). If n = 1 , there exists z of such form satisfying (6.7): z(r) \equiv -1 , consistently with (6.8). On the other hand, if n = 2 , applying (6.7) to (6.2) with \lambda = c_1 = c_2 = 0 leads to a contradiction.
Summing up, all complements of balls are calibrable if n\neq 2 . On the other hand, if n = 2 all complements of balls turn out not to be calibrable.
Let now U = A_{R_0}^{R_1} = B_{R_1}\setminus B_{R_0} , 0 < R_0 < R_1 . In this case \partial U has two connected components, so there exist two distinct choices of signature: constant and non-constant. Let us first consider the former. To focus attention, we choose \chi \equiv -1 . Then, boundary conditions (5.4) take form
\begin{equation} z(R_0) = 1, \quad z(R_1) = -1, \quad z'(R_0) = z'(R_1) = 0. \end{equation} | (6.9) |
Applying (6.9) to (6.2) or (6.3) leads to a system of four affine equations with four unknowns. In the case n \neq 2 , the solution is
\begin{align} \begin{split} c_0 & = \tfrac{- (R_1-R_0) R_1^n R_0^n \left( (n-1)(n-2) (R_1+R_0)^2 - 2 R_0 R_1\right)+2 R_1^3 R_0^{2 n}-2 R_0^3 R_1^{2 n}}{R_1 R_0 \left(R_1^n R_0^n \left(n^2 \left(R_1^2-R_0^2\right)^2+8 R_1^2 R_0^2\right)-4 R_1^2 R_0^{2 n+2}-4 R_0^2 R_1^{2 n+2}\right)}, \\ c_1 & = \tfrac{(R_1+R_0) \left(R_1^n \left(2 (n-1) R_1^2 + (n+2) R_1 R_0 - (n+2) R_0^2 \right) + R_0^n \left((n+2)R_1^2 - (n+2)R_1 R_0 - 2(n-1)R_0^2 \right)\right)}{2 \left(n^2-4\right) R_1^3 R_0^3 + 4 R_1^{3-n} R_0^{n+3} + 4 R_1^{n+3} R_0^{3-n} - n^2 R_1^5 R_0 - n^2 R_1 R_0^5}, \\ c_2 & = \tfrac{(R_1-R_0) R_1^n R_0^n \left(6 R_1^2 R_0^2 + (n-1) n R_1^3 R_0 + (n-1) n R_1^4 + (n-1) n R_1 R_0^3 + (n-1) n R_0^4 \right) - 6 R_1^3 R_0^{2 n+2} + 6 R_0^3 R_1^{2 n+2}}{R_1 R_0 \left(R_1^n R_0^n \left(n^2 \left(R_1^2-R_0^2\right)^2+8 R_1^2 R_0^2\right)-4 R_1^2 R_0^{2 n+2}-4 R_0^2 R_1^{2 n+2}\right)}, \\ c_3 & = \tfrac{(R_1 + R_0) \left(R_0^2 R_1^n \left(-2 (n-3) R_1^2 - n R_1 R_0 + n R_0^2\right)- R_1^2 R_0^n \left(n R_1^2 - R_0 (n R_1 + 2(n-3)R_0)\right)\right)}{2 \left(n^2-4\right) R_1^3 R_0^3 + 4 R_1^{3-n} R_0^{n+3} + 4 R_1^{n+3} R_0^{3-n} - n^2 R_1^5 R_0 - n^2 R_1 R_0^5}. \end{split} \end{align} | (6.10) |
This can be rewritten in a form emphasizing homogeneity:
\begin{align} \begin{split} c_0 & = \tfrac{- (Q-1) Q^n \left((n-1)(n-2) (Q+1)^2 - 2Q\right)+2 Q^3 - 2 Q^{2 n}}{ Q^{n-2} \left(n^2 \left(Q^2-1\right)^2+8 Q^2 \right)-4 - 4 Q^{2 n}} R_1^{-3}, \\ c_1 & = \tfrac{(Q+1) \left(Q^n \left(2 (n-1) Q^2 + (n+2) Q - (n+2) \right) + \left((n+2)Q^2 - (n+2)Q - 2(n-1) \right)\right)}{2 \left(n^2-4\right) Q^n + 4 + 4 Q^{2n} - n^2 Q^{n+2} - n^2 Q^{n-2} } R_1^{n-3}, \\ c_2 & = \tfrac{(Q-1) Q^n \left(6 Q^2 + (n-1) n Q^3 + (n-1) n Q^4 + (n-1) n Q + (n-1) n \right) - 6 Q^3 + 6 Q^{2 n+2}}{Q^n \left(n^2 \left(Q^2-1\right)^2+8 Q^2 \right)-4 Q^2 -4 Q^{2 n+2}} R_1^{-1}, \\ c_3 & = \tfrac{(Q + 1) \left( Q^n \left(-2 (n-3) Q^2 - n Q + n \right)- Q^2 \left(n Q^2 - (n Q + 2(n-3))\right)\right)}{2 \left(n^2-4\right) Q^{n+2} + 4 Q^{2} + 4 Q^{2n+2} - n^2 Q^{n+4} - n^2 Q^n} R_1^{n-1}, \end{split} \end{align} | (6.11) |
where we denoted Q = R_1/R_0 . We can further simplify it to
\begin{align} \begin{split} c_0 & = \tfrac{2Q^3(Q^{2n-3}-1) + (Q-1)Q^n ( (n-1)(n-2)(Q+1)^2 - 2Q )}{4(Q^n-1)^2 - n^2(Q^2-1)^2Q^{n-2}} R_1^{-3}, \\ c_1 & = \tfrac{(Q+1) ( 2(n-1)(Q^{n+2}-1)+(n+2)Q(Q-1)(Q^{n-1}+1) )}{4(Q^n-1)^2 - n^2(Q^2-1)^2Q^{n-2}} R_1^{n-3}, \\ c_2 & = -\tfrac{6Q(Q^{2n-1}-1) + (Q-1)Q^{n-2} ( 6Q^2 + n(n-1)(1+Q)(1+Q^3) )}{4(Q^n-1)^2 - n^2(Q^2-1)^2Q^{n-2}} R_1^{-1}, \\ c_3 & = -\tfrac{(Q+1) ( 2(n-3)(Q^n-1)+nQ(Q-1)(Q^{n-3}+1) )}{4(Q^n-1)^2 - n^2(Q^2-1)^2Q^{n-2}} R_1^{n-1}. \end{split} \end{align} | (6.12) |
We need to check whether condition |Z|\leq 1 is satisfied. We calculate
\begin{eqnarray} z''(r) = 6 c_0 r + (n-3)(n-2) c_1 r^{1-n} + n(n-1) c_3 r^{-n-1}\\ = r^{-n-1} (6 c_0 r^{n+2} + (n-3)(n-2) c_1 r^2 + n(n-1) c_3) = : r^{-n-1} w(r). \end{eqnarray} | (6.13) |
Using the form (6.12), we can check that c_0 > 0 , c_1 > 0 for all Q > 1 . Therefore, w has at most one zero on the half-line r > 0 . Consequently, z'' has at most one zero, so z has at most one inflection point. Taking into account (6.9), z cannot have a local extremum on ]R_0, R_1[. Thus, |z|\leq 1 on ]R_0, R_1[ and Z is a valid calibration.
In the case n = 2 , the solution is
\begin{align*} c_0 & = \tfrac{R_1^2 - R_0^2 + 2 R_1 R_0 \log(R_1/R_0)}{4 R_1 (R_1 - R_0) R_0 (-R_1^2 + R_0^2 + (R_1^2 + R_0^2) \log(R_1/R_0))} \\ c_1 & = \tfrac{-R_1^3 - 3 R_1^2 R_0 - 3 R_1 R_0^2 - R_0^3}{2 R_1 R_0 (-R_1^2 + R_0^2 + (R_1^2 + R_0^2) \log(R_1/R_0))} \\ c_2 & = \tfrac{-3 R_1^4 + 3 R_0^4 + 2 (R_1^4 - R_1^3 R_0 + R_1^2 R_0^2 - 3 R_1 R_0^3) \log(R_1) + 2(3 R_1^3 R_0 - R_1^2 R_0^2 + R_1 R_0^3 - R_0^4) \log(R_0)}{4 R_1 (R_1 - R_0) R_0 (-R_1^2 + R_0^2 + (R_1^2 + R_0^2) \log(R_1/R_0))} \\ c_3 & = \tfrac{R_1 (-3 R_1^2 R_0 + 3 R_0^3 + 2 (R_1^2 R_0 - R_1 R_0^2 + R_0^3) \log(R_1/R_0) )}{4 (R_1 - R_0) (-R_1^2 + R_0^2 + (R_1^2 + R_0^2) \log(R_1/R_0))} \end{align*} |
which can be rewritten (again, denoting Q = R_1/R_0 ) as
\begin{align} \begin{split} c_0 & = \tfrac{Q^2\left(Q^2 - 1 + 2 Q \log Q\right)}{4 (Q-1)\left(-Q^2 + 1 +(Q^2 + 1) \log Q\right)} R_1^{-3}, \\ c_1 & = \tfrac{-(Q+1)^3}{2 \left(-Q^2 + 1 + (Q^2 + 1) \log Q\right)} R_1^{-1}, \\ c_2 & = \tfrac{-3(Q^4 - 1) - 2(3 Q^3 - Q^2 + Q - 1)\log Q}{4(Q-1)\left(-Q^2 + 1 +(Q^2 +1) \log Q\right)}R_1^{-1} + \tfrac{(Q+1)^3}{2 \left(-Q^2 + 1 + (Q^2 + 1) \log Q\right)}R_1^{-1} \log R_1, \\ c_3 & = \tfrac{-3 Q^2 + 3 + 2(Q^2 - Q + 1) \log Q}{4 (Q-1) \left( -Q^2 + 1 +(Q^2 + 1) \log Q \right)}R_1. \end{split} \end{align} | (6.14) |
As before, we calculate the second derivative of z :
\begin{equation*} z''(r) = 6 c_0 r + c_1 r^{-1} +2 c_3 r^{-3} = r^{-3} (6 c_0 r^4 + c_1 r^2 + 2 c_3) = : r^{-3} w(r). \end{equation*} |
The polynomial w has at most 2 positive roots, and so does z'' . By (6.9), at least one of them belongs to ]R_0, R_1[. Furthermore, since c_0 > 0 for Q > 1 , z''(r) is positive for large values of r . Taking into account these observations, we deduce that |z|\leq 1 on [R_0, R_1] if and only if z''(R_0) \leq 0 (compare Figure 1). This inequality is equivalent to
m(Q): = \log Q - \frac{(Q^2 - 1)(2Q - 1)}{Q(Q^2 - 2Q + 3)} \leq 0. |
We compute
\begin{equation} m(1) = 0, \qquad \lim\limits_{Q \to +\infty} m(Q) = +\infty, \qquad m'(Q) = \frac{(Q-3)(Q-1)(Q+1)^3}{Q^2(Q^2- 2Q + 3)^2}. \end{equation} | (6.15) |
We observe that m has exactly one zero Q_* on ]1, +\infty[, and m(Q) \leq 0 if and only if Q \leq Q_* . Therefore, Z is a valid calibration for A_{R_0}^{R_1} with signature -1 if and only if R_1/R_0 \leq Q_* . By (6.15) it is evident that Q_* > 3 . Numerical computation using Wolfram Mathematica shows that Q_* \approx 9.7 . Thus, A_{R_0}^{R_1} with constant signature is calibrable if and only if R_1/R_0 \leq Q_* . This concludes the proof of Theorem 3.
Now, let us consider non-constant signature. We assume that \chi = 1 on \partial B_{R_0} and \chi = -1 on \partial B_{R_1} . This choice leads to
\begin{equation} z(R_0) = -1, \quad z(R_1) = -1, \quad z'(R_0) = z'(R_1) = 0. \end{equation} | (6.16) |
If n\neq 2 , the solution to the resulting affine system is
\begin{align} \begin{split} c_0 & = \tfrac{-2 R_1^3 R_0^{2 n}-2 R_0^3 R_1^{2 n}+R_0^n R_1^n (R_0+R_1) \left((n-2) (n-1) R_0^2-2 ((n-3) n+1) R_0 R_1+(n-2) (n-1) R_1^2\right)}{R_0 R_1 \left(R_0^n R_1^n \left(n^2 \left(R_0^2-R_1^2\right)^2+8 R_0^2 R_1^2\right)-4 R_1^2 R_0^{2 n+2}-4 R_0^2 R_1^{2 n+2}\right)} \\ c_1 & = -\tfrac{-3 n R_1 R_0^{n+2}+2 (n-1) R_0^{n+3}+(n+2) R_1^3 R_0^n+(n+2) R_0^3 R_1^n-3 n R_0 R_1^{n+2}+2 (n-1) R_1^{n+3}}{4 R_0^{3-n} R_1^{3-n} \left(R_0^n-R_1^n\right)^2-n^2 R_0 R_1 \left(R_0^2-R_1^2\right)^2} \\ c_2 & = \tfrac{6 R_1^3 R_0^{2 n+2}+6 R_0^3 R_1^{2 n+2}-R_0^n R_1^n (R_0+R_1) \left((n-1) n R_0^4-(n-1) n R_0^3 R_1-(n-1) n R_0 R_1^3+(n-1) n R_1^4+6 R_0^2 R_1^2\right)}{R_0 R_1 \left(R_0^n R_1^n \left(n^2 \left(R_0^2-R_1^2\right)^2+8 R_0^2 R_1^2\right)-4 R_1^2 R_0^{2 n+2}-4 R_0^2 R_1^{2 n+2}\right)} \\ c_3 & = \tfrac{(R_0-R_1) \left(R_0^2 R_1^n \left(n (R_0-R_1) (R_0+2 R_1)+6 R_1^2\right)-R_1^2 R_0^n \left(-2 (n-3) R_0^2+n R_0 R_1+n R_1^2\right)\right)}{4 R_0^{3-n} R_1^{3-n} \left(R_0^n-R_1^n\right)^2-n^2 R_0 R_1 \left(R_0^2-R_1^2\right)^2} \end{split} \end{align} | (6.17) |
which we rewrite as
\begin{align} \begin{split} c_0 & = \tfrac{-2 Q^3-2 Q^{2 n}+ Q^n (1+Q) \left((n-2) (n-1) -2 ((n-3) n+1) Q+(n-2) (n-1) Q^2\right)}{Q^{n-2} \left(n^2 \left(1-Q^2\right)^2+8 Q^2\right)-4 -4 Q^{2 n}} R_1^{-3} \\ c_1 & = -\tfrac{-3 n Q+2 (n-1) +(n+2) Q^3 +(n+2) Q^n-3 n Q^{n+2}+2 (n-1) Q^{n+3}}{4 \left(1-Q^n\right)^2-n^2 Q^{n-2} \left(1-Q^2\right)^2} R_1^{n-3}\\ c_2 & = \tfrac{6 Q^3 +6 Q^{2 n+2}-Q^n (1+Q) \left((n-1) n -(n-1) n Q-(n-1) n Q^3+(n-1) n Q^4+6 Q^2\right)}{ Q^n \left(n^2 \left(1-Q^2\right)^2+8 Q^2\right)-4 Q^2 -4 Q^{2 n+2}} R_1^{-1}\\ c_3 & = \tfrac{(1-Q) \left( Q^n \left(n (1-Q) (1+2 Q)+6 Q^2\right)-Q^2 \left(-2 (n-3) +n Q+n Q^2\right)\right)}{4 Q^2 \left(1-Q^n\right)^2-n^2 Q^n \left(1-Q^2\right)^2} R_1^{n-1} \end{split} \end{align} | (6.18) |
We note that in case n = 1 the solution reduces to
\begin{equation*} c_0 = 0, \quad c_1 = 0, \quad c_2 = 0, \quad c_3 = -1, \end{equation*} |
while in case n = 3 it reduces to
\begin{equation*} c_0 = 0, \quad c_1 = -1, \quad c_2 = 0, \quad c_3 = 0. \end{equation*} |
In both of these cases z is constant and we have \lambda = c_0 = 0 . On the other hand, if n \geq 4 , we can check that c_0 > 0 for Q > 1 . Recalling (6.13), we observe that z'' has at most two zeros on the positive half-line and z''(r) > 0 for large values of r . On the other hand, by (6.16), if z has N local extrema on ]R_0, R_1[, it needs to have at least N+1 inflection points. We deduce from these conditions that z has exactly one local maximum and no local minima, and therefore z \geq -1 on ]R_0, R_1[ (compare Figure 2). It remains to check whether z\leq 1 on ]R_0, R_1[. Let now
f(r) = r^{1-n}(r^{n-1}z(r))' = f'(r) + (n-1)\frac{f(r)}{r}. |
Then, by (6.1), (6.16), f is a solution to the second-order elliptic problem
\mathcal A f = \lambda, \quad f(R_0) = -\frac{n-1}{R_0}, \quad f(R_1) = -\frac{n-1}{R_1}, |
where
\mathcal A f = - r^{1-n}(r^{n-1}f'(r))' = - f''(r) - (n-1)\frac{f'(r)}{r}. |
Since c_0 \geq 0 , we have \lambda < 0 for Q > 1 . By the classical weak maximum principle [19, Theorem 3.1.],
\max\limits_{[R_0, R_1]}f = \max\limits_{\{R_0, R_1\}}f = -\frac{n-1}{R_1}. |
Now, if z has a local maximum at r_0 , then z'(r_0) = 0 , so f(r_0) = \frac{z(r_0)}{r_0} . Consequently,
\frac{z(r_0)}{r_0} \leq -\frac{n-1}{R_1} < 0, |
so z < 0 on [R_0, R_1] . Thus, if n\neq 2 , all annuli with non-constant signature are calibrable.
We move to the case n = 2 . Now, the solution to the affine system for coefficients of z is
\begin{align*} \begin{split} c_0 & = \tfrac{-R_0^2 - 2 R_0 R_1 \log (R_1/R_0)+R_1^2}{4 R_0 R_1 (R_0+R_1) \left(-R_0^2+R_1^2- \left(R_0^2+R_1^2\right)\log (R_1/R_0)\right)}, \\ c_1 & = \tfrac{(R_0-R_1)^3}{2 R_0 R_1 \left(-R_0^2+R_1^2- \left(R_0^2+R_1^2\right)\log (R_1/R_0)\right)}, \\ c_2 & = \tfrac{(R_0-R_1) (R_0+R_1) \left(R_0^2+4 R_0 R_1+R_1^2\right)-2 R_0 \left(R_0^3+R_0^2 R_1+R_0 R_1^2+3 R_1^3\right) \log (R_0)+2 R_1 \left(3 R_0^3+R_0^2 R_1+R_0 R_1^2+R_1^3\right) \log (R_1)}{4 R_0 R_1 (R_0+R_1) \left(-R_0^2+R_1^2- \left(R_0^2+R_1^2\right)\log (R_1/R_0)\right)}, \\ c_3 & = \tfrac{R_0 R_1 \left(2 \left(R_0^2+R_0 R_1+R_1^2\right) \log (R_1/R_0))+3 (R_0-R_1) (R_0+R_1)\right)}{4 (R_0+R_1) \left(-R_0^2+R_1^2- \left(R_0^2+R_1^2\right)\log (R_1/R_0)\right)} \end{split} \end{align*} |
or equivalently
\begin{align*} \begin{split} c_0 & = \tfrac{Q^2\left(-1 - 2 Q \log Q+Q^2\right)}{4 (1+Q) \left(-1 + Q^2- \left(1+Q^2\right)\log Q \right)}R_1^{-3}, \\ c_1 & = \tfrac{(1-Q)^3}{2 \left(-1 + Q^2- \left(1 + Q^2\right)\log Q\right)}R_1^{-1}, \\ c_2 & = \tfrac{(1-Q) (1+Q) \left(1 + 4 Q + Q^2\right) + 2 \left(1 + Q + Q^2 + 3 Q^3\right) \log Q}{4 (1 + Q) \left(-1 + Q^2 - \left(1 + Q^2\right)\log Q \right)}R_1^{-1} + \tfrac{ (Q-1)^3}{2 \left(-1 + Q^2 - \left(1 + Q^2\right)\log Q \right)}R_1^{-1}\log R_1, \\ c_3 & = \tfrac{2 \left(1 + Q + Q^2\right) \log Q +3 (1 - Q)(1 + Q)}{4(1 + Q) \left(-1 + Q^2 - \left(1 + Q^2\right)\log Q\right)}R_1. \end{split} \end{align*} |
We can check that in this case c_0 < 0 for Q > 1 . By the same argument as in the previous case, we show that z < -1 in ]R_0, R_1[ (compare Figure 2), so it does not define a valid calibration. Thus, in the case n = 2 all annuli with non-constant signature are not calibrable.
In this section, our goal is to provide explicit description of solutions to (1.1) emanating from the characteristic function of a ball
\begin{equation} u_0 = a_0 \mathbf 1_{B_{R_0}}. \end{equation} | (7.1) |
In the case of second-order total variation flow, the solutions with initial datum (7.1) are known to be of form
u(t) = a(t)\mathbf 1_{B_{R_0}} |
with finite extinction time, i.e., there exists t_* > 0 such that a(t) = 0 for t \geq t_* . In the fourth order case, based on the treatment of case n = 1 in [12], we would expect the solutions to have the form
\begin{equation} u(t) = a(t)\mathbf 1_{B_{R(t)}}, \end{equation} | (7.2) |
at least until an extinction time beyond which u(t, \cdot) \equiv 0 . This intuition turns out to be correct in every dimension except n = 2 .
Let first n \geq 3 . As we have checked in Section 6, in this case both balls and complements of balls are calibrable. Thus, as long as the solution is of form (7.2) in time instance t \geq 0 , we expect a valid Cahn-Hoffman vector field Z to be given by
\begin{equation} Z(x) = \left\{\begin{array}{l} Z_{in}(x) \text{ if } |x| \in [0, R[ \\ Z_{out}(x) \text{ if } |x| > R, \end{array} \right. \end{equation} | (7.3) |
where Z_{in} is the calibration w constructed for a ball B_R and Z_{out} is the calibration we constructed for the complement of that ball, recall:
\begin{equation} \lambda = - \frac{n(n+2)}{R^3}, \end{equation} | (7.4) |
\begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{|x|}{R}\right)^3 \frac{x}{|x|} -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \frac{n-1}{2} \left(\frac{|x|}{R}\right)^{3-n}\frac{x}{|x|} + \frac{n-3}{2} \left(\frac{|x|}{R}\right)^{1-n}\frac{x}{|x|}. \end{equation} | (7.5) |
We further calculate:
\begin{equation*} \operatorname{div} Z_{in}(x) = \frac{n+2}{2} \frac{|x|^2}{R^3} - \frac{3n}{2} \frac{1}{R}, \qquad \operatorname{div} Z_{out}(x) = - (n-1) \frac{|x|^{2-n}}{R^{3-n}}, \end{equation*} |
\begin{equation*} \nabla \operatorname{div} Z_{in}(x) = (n+2) \frac{x}{R^3}, \qquad \nabla \operatorname{div} Z_{out}(x) = (n-1)(n-2) \frac{|x|^{-n} x}{R^{3-n}}. \end{equation*} |
It is straightforward to check that \operatorname{div} Z \in D^1(\mathbb{R}^n) \cap L^{2^*}(\mathbb{R}^n) = D_0^1(\mathbb{R}^n) . Next, we deduce
\begin{eqnarray} u_t = -\Delta \operatorname{div} Z \\ = - \Delta \operatorname{div} Z_{in}\, \mathcal{L}^n \, ∟\, _{B_R} - \Delta \operatorname{div} Z_{out}\, \mathcal{L}^n \, ∟\, _{\mathbb{R}^n\setminus B_R} + \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z_{in} - \nabla \operatorname{div} Z_{out})\, \mathcal{H}^{n-1} \, ∟\, _{\partial B_R} \\ = - \frac{n(n+2)}{R^3}\, \mathcal{L}^n \, ∟\, _{B_R} - \frac{n(n-4)}{R^2}\, \mathcal{H}^{n-1} \, ∟\, _{\partial B_R}. \end{eqnarray} | (7.6) |
Then, using the identity \frac{\mathrm{d}}{\mathrm{d} t} \int_{\mathbb{R}^n} u = \int_{\mathbb{R}^n} u_t , we obtain (recall notation (7.2))
\begin{equation*} a(t) \mathcal{H}^{n-1}\left(\partial B_{R(t)}\right) \frac{\mathrm{d} R}{\mathrm{d} t} = a(t) \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{L}^n\left(B_{R(t)}\right) = - \frac{n(n-4)}{R^2} \mathcal{H}^{n-1}\left(\partial B_{R(t)}\right). \end{equation*} |
Summing up, evolution of initial datum (7.1) is given by (7.2) with a , R satisfying
\begin{equation} \frac{\mathrm{d} a}{\mathrm{d} t} = - \frac{n(n+2)}{R^3}, \qquad \frac{\mathrm{d} R}{\mathrm{d} t} = - \frac{n(n-4)}{R^2 a}. \end{equation} | (7.7) |
This system can be explicitly solved by noticing that
\begin{equation*} \frac{\mathrm{d}}{\mathrm{d} t} (a R^3) = -n(n+2) - 3n(n-4) = -n (4n - 10) \end{equation*} |
and therefore
\begin{equation*} a R^3 = a_0 R_0^3 - n (4n - 10) t \end{equation*} |
along trajectories. The solution is
\begin{equation} a(t) = a_0 \left(1 - \frac{n(4n-10)}{a_0 R_0^3} t\right)^{\frac{n+2}{4n-10}}, \qquad R(t) = R_0 \left(1 - \frac{n(4n-10)}{a_0 R_0^3} t\right)^{\frac{n-4}{4n-10}}. \end{equation} | (7.8) |
We note that the solution satisfies
\left(\frac{a}{a_0}\right)^{n-4} = \left(\frac{R}{R_0}\right)^{n+2} |
along trajectories. (This "first integral" could also have been used to solve the system (7.7).) Let us point out a few observations concerning the solutions (compare Figure 3):
● the extinction time is equal to t_* = \frac{a_0 R_0^3}{n(4n-10)} ,
● if n = 3 , R(t) is increasing and R(t) \to + \infty as t \to t_*^- ,
● if n = 4 , R(t) = R_0 is constant,
● in higher dimensions, R(t) is decreasing and R(t) \to 0 as t \to t_*^- .
In the case n = 2 we were able to exhibit a calibration for the ball B_R , but not for its complement. Another possible ansatz on the Cahn-Hoffman vector field of form (7.3) is one where Z_{in} is the calibration we constructed for B_R and Z_{out} is the choice considered in [15]:
\begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{|x|}{R}\right)^3 \frac{x}{|x|} -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \frac{x}{|x|}. \end{equation} | (7.9) |
We calculate
\begin{equation} \operatorname{div} Z_{out} = - \frac{(n-1)}{|x|}, \qquad \nabla \operatorname{div} Z_{out} = \frac{(n-1)x}{|x|^3}, \end{equation} | (7.10) |
hence
\begin{equation} u_t(t, x) = - \frac{(n-1)(n-3)}{|x|^3} \quad \text{in } \mathcal{D}'\left(\mathbb{R}^n \setminus \overline{B}_{R(t)}\right). \end{equation} | (7.11) |
If n \geq 4 , this would lead to u(t) being radially strictly increasing for positive t and large values of |x| , which would be at odds with our choice of Z_{out} . In fact, if n \geq 4 , \operatorname{div} Z \not \in D_0^1(\mathbb{R}^n) for any Z of this form. However, in smaller dimensions this ansatz remains a viable option. If n = 3 , it leads to the same solution as before. On the other hand, if n = 2 , we obtain a solution which is not of form (7.2). Instead, we are led to assume
\begin{equation} u(t, x) = a(t) \mathbf 1_{B_{R(t)}} + \frac{t}{|x|^3}\mathbf 1_{\mathbb{R}^2\setminus B_{R(t)}}. \end{equation} | (7.12) |
We have:
\begin{equation*} \operatorname{div} Z_{in}(x) = 2 \frac{|x|^2}{R^3} - 3 \frac{1}{R}, \qquad \operatorname{div} Z_{out}(x) = - \frac{1}{|x|}, \end{equation*} |
\begin{equation*} \nabla \operatorname{div} Z_{in}(x) = 4 \frac{x}{R^3}, \qquad \nabla \operatorname{div} Z_{out}(x) = \frac{x}{|x|^3}, \end{equation*} |
\begin{equation*} u_t = -\Delta \operatorname{div} Z = - \frac{8}{R^3}\, \mathcal{L}^2 \, ∟\, _{B_R} + \frac{1}{|x|^3}\, \mathcal{L}^2 \, ∟\, _{\mathbb{R}^2 \setminus B_R} + \frac{3}{R^2}\, \mathcal{H}^1 \, ∟\, _{\partial B_R} \end{equation*} |
and, recalling (7.12),
\begin{equation*} \left(a(t) - \frac{t}{R(t)^3}\right) \mathcal{H}^1(\partial B_{R(t)}) \frac{\mathrm{d} R}{\mathrm{d} t} = \left(a(t) - \frac{t}{R(t)^3}\right) \frac{\mathrm{d} }{\mathrm{d} t} \mathcal{L}^2(B_{R(t)}) = \frac{3}{R(t)^2} \mathcal{H}^1(\partial B_{R(t)}). \end{equation*} |
Thus, we arrive at ODE system
\begin{equation} \frac{\mathrm{d} a}{\mathrm{d} t} = -\frac{8}{R^3}, \qquad \frac{\mathrm{d} R}{\mathrm{d} t} = \frac{3R}{aR^3 - t}. \end{equation} | (7.13) |
This system is not autonomous, but it can be integrated by noticing that along trajectories
\begin{equation*} \frac{\mathrm{d}}{\mathrm{d} t} \left(aR^3 - t\right) = \frac{9aR^3}{aR^3 - t} - 8 - 1 = \frac{9t}{aR^3 - t} \end{equation*} |
and so
\begin{equation*} aR^3 = \sqrt{a_0^2 R_0^6 + 9 t^2} + t. \end{equation*} |
This implies, first of all, that
\begin{equation*} a(t) > \frac{t}{R(t)^3} \end{equation*} |
for all t > 0 and the form of solution (7.12) is preserved as long as the solution does not vanish. Furthermore, we can rewrite the system (7.13) in decoupled form
\begin{equation} \frac{\mathrm{d}}{\mathrm{d} t} \log a = \frac{- 8}{\sqrt{a_0^2 R_0^6 + 9 t^2} + t}, \qquad \frac{\mathrm{d}}{\mathrm{d} t} \log R^3 = \frac{9}{\sqrt{a_0^2 R_0^6 + 9 t^2}}. \end{equation} | (7.14) |
These equations can be explicitly integrated:
\begin{equation*} a(t) = a_0 \frac{a_0^4 R_0^{12} + 8 a_0^2 R_0^{12}t^2}{\left(\sqrt{a_0^2 R_0^6 + 9 t^2} + 3 t \right)^2 \left(a_0^2 R_0^6 + 6 t^2 + 2 t \sqrt{a_0^2 R_0^6 + 9 t^2}\right)}, \end{equation*} |
\begin{equation*} R(t) = R_0 \sqrt{1 + 6t \frac{3 t + \sqrt{a_0^2 R_0^6 + 9 t^2}}{a_0^2 R_0^6}}. \end{equation*} |
We observe that the solutions exist globally and
\begin{equation*} \lim\limits_{t \to \infty} a(t) = 0, \qquad \lim\limits_{t \to \infty} R(t) = \infty. \end{equation*} |
In particular u stays in the form (7.12) for all t > 0 .
Finally we consider n = 1 . In this case, both ansätze considered before lead to the same solution:
\begin{equation} Z_{in}(x) = \frac{1}{2} \left(\frac{x}{R}\right)^3 -\frac{3}{2} \frac{x}{R}, \qquad Z_{out}(x) = - \mathrm{sgn}\, x \end{equation} | (7.15) |
which coincides with (7.5). Repeating the calculations following (7.5), we obtain a solution of form (7.2) satisfying (7.8), i.e.,
\begin{equation} u(t) = a(t)\mathbf 1_{B_{R(t)}}, \qquad a(t) = a_0 \left(1 + \frac{6}{a_0 R_0^3} t\right)^{-\frac{1}{2}}, \qquad R(t) = R_0 \left(1 + \frac{6}{a_0 R_0^3} t\right)^{\frac{1}{2}}. \end{equation} | (7.16) |
Note that now, as opposed to the case n \geq 3 , the coefficient multiplying t is positive. Like in n = 2 , the extinction time is infinite and we have
\begin{equation*} \lim\limits_{t \to \infty} a(t) = 0, \qquad \lim\limits_{t \to \infty} R(t) = \infty. \end{equation*} |
This concludes the proof of Theorem 4.
Using the calibrations we constructed for generalized annuli, we will now derive a system of ODEs locally prescribing the solution emanating from any piecewise constant, radially symmetric datum (a stack).
Definition 32. Let w \in D(TV) . We say that w is a stack if there exists a number N \in \mathbb{N} and sequences 0 < R^0 < R^1 < \ldots < R^{N-1} , a^0, a^1, \ldots, a^N with a^k \in \mathbb{R} such that
w = a^0 \mathbf{1}_{B_{R^0}} + a^1 \mathbf{1}_{A_{R^0}^{R^1}} +\ldots + a^{N-1} \mathbf{1}_{A_{R^{N-2}}^{R^{N-1}}} + a^N \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}}}. |
Suppose first that n\neq 2 , in which case all connected components of level sets of any stack w are calibrable. Let u_0 be a stack
\begin{equation} u_0 = a^0_0 \mathbf{1}_{B_{R^0_0}} + a^1_0 \mathbf{1}_{A_{R^0_0}^{R^1_0}} +\ldots + a^{N-1}_0 \mathbf{1}_{A_{R^{N-2}_0}^{R^{N-1}_0}} + a^N_0 \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}_0}}, \end{equation} | (7.17) |
where a^{k-1}\neq a^k for k = 1, \ldots, N , a^N_0 = 0 . We expect that if u is the solution emanating from u_0 , then u(t, \cdot) is a stack of form
\begin{equation} u(t, \cdot) = a^0(t) \mathbf{1}_{B_{R^0(t)}} + a^1(t) \mathbf{1}_{A_{R^0(t)}^{R^1(t)}} +\ldots + a^{N-1}(t) \mathbf{1}_{A_{R^{N-2}(t)}^{R^{N-1}(t)}} + a^N(t) \mathbf{1}_{\mathbb{R}^n \setminus B_{R^{N-1}(t)}}, \end{equation} | (7.18) |
with a^N(t) = 0 for all t > 0 , and that a^{k-1}\neq a^k , k = 1, \ldots, N for small t . We construct a Cahn-Hoffman vector field Z(t, \cdot) for u(t, \cdot) by pasting together calibrations Z^k for B_{R^0(t)} , A_{R^k(t)}^{R^{k+1}(t)} , \mathbb{R}^n \setminus B_{R^{N-1}(t)} with suitable choice of signatures. We have
\begin{eqnarray} u_t = -\Delta \operatorname{div} Z = - \Delta \operatorname{div} Z^0\, \mathcal{L}^n \, ∟\, _{B_{R^0}} - \sum\limits_{k = 1}^{n} \Delta \operatorname{div} Z^k\, \mathcal{L}^n \, ∟\, _{A_{R^{k-1}}^{R^k}} - \Delta \operatorname{div} Z^N\, \mathcal{L}^n \, ∟\, _{\mathbb{R}^n\setminus B_{R^N}} \\ + \sum\limits_{k = 0}^{n} \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z^k - \nabla \operatorname{div} Z^{k+1})\, \mathcal{H}^{n-1} \, ∟\, _{S_{R^k}}. \end{eqnarray} | (7.19) |
We denote
\begin{eqnarray} \left. \frac{x}{|x|} \cdot(\nabla \operatorname{div} Z^k - \nabla \operatorname{div} Z^{k+1}) \right|_{S_{R^k}} \\ = z^k_{rr}(R^k) - z^{k+1}_{rr}(R^k) + \frac{n-1}{R^k}(z^k_{r}(R^k)- z^{k+1}_{r}(R^k))-\frac{n-1}{(R^k)^2}(z^k(R^k)- z^{k+1}(R^k))\\ = z^k_{rr}(R^k) - z^{k+1}_{rr}(R^k) = : d^k. \end{eqnarray} |
The values of d^k are functions of R^0, \ldots, R^{N-1} . Assuming that R^k are regular enough and \varepsilon , |t-s| are small enough, we have
\frac{\mathrm{d}}{\mathrm{d} t}\int_{A_{R^k(s)-\varepsilon}^{R^k(s)+\varepsilon}} u = \int_{A_{R^k(s)-\varepsilon}^{R^k(s)+\varepsilon}} u_t, |
whence
(a^k(t) - a^{k-1}(t)) \mathcal{H}^{n-1}(S_{R^k(t)}) \frac{\mathrm{d} R^k}{\mathrm{d} t} = (a^k(t) - a^{k-1}(t)) \frac{\mathrm{d}}{\mathrm{d} t} \mathcal{L}^n(A_{R^k(s)-\varepsilon}^{R^k(t)}) = d^k(t) \mathcal{H}^{n-1}(S_{R^k(t)}). |
Further, for k = 0, \ldots, N , we denote by \lambda^k the value of -\Delta \operatorname{div} Z^k(t, \cdot) which is constant since Z^k is a calibration. Then, we can write down the system of ODEs for a^k and R^k :
\begin{equation} \frac{\mathrm{d} a^k}{\mathrm{d} t} = \lambda^k \text{ for } k = 0, \ldots N, \qquad \frac{\mathrm{d} R^k}{\mathrm{d} t} = \frac{d^k}{a^k - a^{k+1}} \text{ for } k = 0, \ldots N-1. \end{equation} | (7.20) |
Let c_0^k denote c_0 given by (6.17) if \mathrm{sgn}\, (a^{k+1} - a^k) = \mathrm{sgn}\, (a^k - a^{k-1}) or by (6.10) if \mathrm{sgn}\, (a^{k+1} - a^k) \neq \mathrm{sgn}\, (a^k - a^{k-1}) , with R^{k+1} and R^k in place of R_1 and R_0 . Then, we have
\begin{equation} \lambda^0 = \mathrm{sgn}\, (a^1-a^0)\frac{n(n+2)}{(R^0)^3}, \quad \lambda^k = 2n(n+2)\mathrm{sgn}\, (a^{k+1} - a^k) c_0^k \text{ for } k = 1, \ldots, N-1, \quad \lambda^N = 0. \end{equation} | (7.21) |
We observe that in a neighborhood of any initial datum R^0_0, \ldots, R^{N-1}_0 , a^0_0, \ldots, a^N_0 , R^k_0 < R^{k+1}_0 , a^k_0 \neq a^{k+1}_0 , the r. h. s. of (7.20) is regular in R^0, \ldots, R^{N-1} , a^0, \ldots, a^N , so locally the system has a unique solution. Unique solvability fails when a time instance t > 0 is reached such that a^k(t) = a^{k+1}(t) , R^k(t) = R^{k+1}(t) or R_0 = 0 . In such case u(t, \cdot) is again a stack with a smaller N , and we can restart our procedure. This concludes the proof of Theorem 5.
Next we deal with the remaining case of dimension n = 2 . In this case, our attempt to obtain a radial calibration failed for complements of balls and for some annuli. Again, let u_0 be a stack of form (7.17). For k = 1, \ldots, N , let \sigma^k = \mathrm{sgn}\, (a^k_0 - a^{k-1}_0) . We assume the following ansatz on the solution u and the associated field Z for small t > 0 :
\begin{equation} u(t, \cdot) = a^0(t) \text{ on } B_{R^0(t)}, u(t, \cdot) = a^k(t) \text{ on } A_{\max(R^{k-1}(t), R^k(t)/Q_*)}^{R^k(t)} \text{ if } \sigma^{k+1} \neq \sigma^k, \ k = 1, \ldots, {N-1}, \end{equation} | (7.22) |
\begin{eqnarray} Z(t, x) = \sigma^k \tfrac{x}{|x|} \text{ on } A_{R^{k-1}(t)}^{R^k(t)} \text{ if } \sigma^{k+1} \neq \sigma^k \text{ or on } A_{R^{k-1}(t)}^{R^k(t)/Q_*} \text{ if }\sigma^{k+1} = \sigma^k, \ k = 1, \ldots, {N-1}, \\ Z(t, x) = \sigma^{k+1} \tfrac{x}{|x|} \text{ on } \mathbb{R}^n \setminus B_{R^{N-1}(t)}. \end{eqnarray} | (7.23) |
We complete the definition of a Cahn-Hoffman field Z consistent with (7.22), (7.23) by pasting the calibrations Z^k with suitable choice of signatures into the gaps left in (7.23). This leads to
u_t(t, \cdot) = \lambda^0(t) \text{ in } \mathcal{D}'(B_{R^0(t)}), |
\begin{align*} u_t(t, x) & = \lambda^k(t) \text{ in } \mathcal{D}'\left(A_{\max(R^{k-1}(t), R^k(t)/Q_*)}^{R_k(t)}\right), \\ u_t(t, x) & = \frac{\sigma^k}{|x|^3} \text{ in } \mathcal{D}'\left(A_{R^{k-1}(t)}^{R^k(t)/Q_*}\right) \text{ if } \sigma^k \neq \sigma^{k+1} \text{ or in } \mathcal{D}'\left(A_{R^{k-1}(t)}^{R^k(t)}\right) \text{ if } \sigma^k = \sigma^{k+1}, \ k = 1, \ldots, N-1, \\ u_t(t, x) & = \frac{\sigma^N}{|x|^3} \text{ in } \mathcal{D}'(\mathbb{R}^2 \setminus B_{R^N(t)}). \end{align*} |
Moreover, u_t(t, \cdot) \in M(\mathbb{R}^2) and
u_t \, ∟\, _{S_{R^k}} = \frac{x}{|x|}\cdot ((\nabla \operatorname{div} Z)^- - (\nabla \operatorname{div} Z)^+) \mathcal{H}^{1} \, ∟\, _{S_{R^k}} = : d^k |
for k = 0, \ldots, N-1 , where (\nabla \operatorname{div} Z)^\pm are the one sided limits as |x| \to (R^k)^\pm . The values of d^k are functions of R^0, \ldots, R^{N-1} . Reasoning as in the case n \neq 2 , the evolution of R^k is governed by equations
\begin{equation} \frac{\mathrm{d} R^k}{\mathrm{d} t} = \frac{d^k}{u(t, x)\big|_{|x| = (R^k)^-} - u(t, x)\big|_{|x| = (R^k)^+}}. \end{equation} | (7.24) |
The values u(t, x)\big|_{|x| = (R^k)^+} are either prescribed by ODEs
\begin{equation} \frac{\mathrm{d} a^k}{\mathrm{d} t} = \lambda^k \end{equation} | (7.25) |
with \lambda^k functions of R^0, \ldots, R^{N-1} in calibrable regions where u(t, x) = a^k(t) , or explicitly determined by u_t(t, x) = \sigma^k /|x|^3 in bending regions. It is important to note that in the case \sigma^k \neq \sigma^{k+1} , R^{k-1} \leq R^k/Q_* the functions d^k, \lambda^k do not depend on R^{k-1} . Thus, one can first solve a part of the system (7.24), (7.25) for the outer annuli, then calculate u in the bending region (without knowing a priori its inner boundary) and move on to solving innermore parts of (7.24), (7.25). This way, finding the solution is indeed again reduced to solving a system of ODEs. We include Figure 4 illustrating the evolution of stacks on the example of the characteristic function of an annulus.
In the case of thick annuli in n = 2 , the qualitative behavior resulting from this procedure is rather intricate and may be surprising. To showcase this, we include Figures 5 and 6 depicting the evolution emanating from the characteristic function of a thick annulus.
Let us explain the evolution in a few words. The inner part of the initial facet corresponding to the annulus instantaneously bends downwards. Meanwhile, the outer boundary of the facet expands outwards, at relatively low speed (practically invisible in the picture). Since the ratio of outer to inner radius of the facet is constant, this means that the whole facet slowly moves outwards. The combined effect of this and the bending results in the very steep (but continuous!) part of the graph between the facet and the bending part. At the same time, the facet corresponding to the inside ball also expands outwards, gradually consuming the bending part. In the final pictured time instance, the whole bending part has disappeared and the solution is a stack again.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors would like to thank the anonymous reviewers for their helpful comments.
The first author of the work was partly supported by the Japan Society for the Promotion of Science (JSPS) through the grants Kakenhi: No. 19H00639, No. 18H05323, No. 17H01091, and by Arithmer Inc. and Daikin Industries, Ltd. through collaborative grants.
The second author of the work was partly supported by JSPS through the grant Kakenhi No. 18H05323.
This work was created during the last author's JSPS Postdoctoral Research Fellowship at the University of Tokyo. The last author was partly supported by the Kakenhi Grant-in-Aid No. 21F20811.
The authors declare no conflict of interest.
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